317823436-Solution-Refrigeration-Air-Conditioning-Stoecker-Jones

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CHAPTER 2 - THERMAL PRINCIPLES 
Page 1 of 5 
2-1. Water at 120 C and a pressure of 250 kPa passes through a pressure-reducing valve and then flows to a separating 
tank at standard atmospheric pressure of 101.3 kPa, as shown in Fig. 2-14. 
(a) What is the state of the water entering the valve (subcooled liquid, saturated liquid, or vapor)? 
(b) For each kilogram that enters the pressure-reducing valve, how much leaves the separating tank as vapor? 
 
 
 
 
Solution: 
 
(a) From Fig. 2-2, a temperature of 120 C and pressure of 250 kPa water lies in the sub-cooled regiom. so it is a sub-
cooled liquid. 
 
(b) At 120 C, h
1
 = 503.72 kJ/kg from Table A-1 
 
 For Pressuring Reducing Valve Dh = 0 
 h
2
 = h
1
 
 
 At 101.3 kPa, Table A-1, h
f
 = 419.06 kJ/kg 
 h
g
 = 2676 kJ/kg 
 
 Let x be the amount of vapor leaving the separating tank. 
 
 h = h
f
 + x(h
g
 - h
f
) 
 
 
419.062676
419.06503.72
hh
hh
x
fg
f
−
−
=
−
−
=
 
 
 x = 0.0375 kg/kg - - - Ans. 
 
2-2. Air flowing at a rate of 2.5 kg/s is heated in a heat exchanger from -10 to 30 C. What is the rate of heat transfer? 
 
Solution: 
 
q = mc
p
(t
2
 - t
1
) 
m = 2.5 kg/s 
c
p
 = 1.0 kJ/kg.K 
t
2
 = 30 C 
t
1
 = -10 C 
 
CHAPTER 2 - THERMAL PRINCIPLES 
Page 2 of 5 
Then, 
 
q = (2.5)(1.0)(30 + 10) 
 
q = 100 kw - - - Ans. 
 
 
2-3. One instrument for measuring the rate of airflow is a venturi, as shown in Fig. 2-15, where the cross-sectional area is 
reduced and the pressure difference between position A and B measured. The flow rate of air having a density of 1.15 
kg/m
3
 is to be measured in a venturi where the area of position A is 0.5 m
2
 and the area at b is 0.4 m
2
. The deflection 
of water (density = 1000 kg/m3) in a manometer is 20 mm. The flow between A and B can be considered to be 
frictionless so that Bernoulli’s equation applies. 
 (a) What is the pressure difference between position A and B? 
 (b) What is the airflow rate? 
 
 
 
 
 
Solution: 
 
(a) Bernoulli equation for manometer 
 
 
B
B
A
A gz
p
gz
p
+
ρ
=+
ρ 
 p
A
 - p
B
 = ρg(z
B
 -z
A
) 
 
 z
B
 - z
A
 = 20 mm = 0.020 m 
 g = 9.81 m/s
2
 
 ρ = 1000 kg/m3 
 
 p
A
 - p
B
 = (1000 kg/m
3
)(9.81 m/s
2
)(0.020 m) 
 
 p
A
 - p
B
 = 196.2 Pa - - - Ans. 
 
(b) Bernoulli Equation for Venturi 
 
 
constant
2
Vp 2
=+
ρ
 
CHAPTER 2 - THERMAL PRINCIPLES 
Page 3 of 5 
 
2
Vp
2
Vp
2
BB
2
AA +
ρ
=+
ρ
 
 
( )2A2B21BA VVpp −ρ=− 
 
 But m = ρA
A
V
A
 = ρA
B
V
B
 
 
 A
A
V
A
 = A
B
V
B
 
 
 AA = 0.5 m2 ans AB = 0.4 m2 
 
 Then 
 
 0.5V
A
 = 0.4V
B
 
 V
A
 = 0.8V
B
 
 
 
( ) ( )[ ]2B2B321BA 0.8VVkg/m1.15Pa196.2pp −==− 
 V
B
 = 30.787 m/s 
 
 Air Flow Rate = A
B
V
B
 
 = (0.4 m
2
)(30.787 m/s) 
 = 12.32 m
3
/s - - - Ans. 
 
2-4. Use the perfect-gas equation with R = 462 J/kg.K to compute the specific volume of saturated vapor at 20 C. Compare 
with data of Table A-1. 
 
Solution: 
 Perfect-Gas Equation: 
 
RTp =ν
 
 
p
RT
=ν
 
 At 20 C, Table A-1, Saturation Pressure = 2.337 kPa = 2337 Pa. 
 Specific volume of saturated vapor = 57.84 m
3
/kg 
 T = 20 C + 273 = 293 K 
 
( )( )
Pa2337
K293J/kg.K462
=ν
 
 
/kgm57.923 3=ν
 
 
 
( )100%
57.84
57.8457.923
Deviation
−
=
 
 Deviation = 0.1435 % 
 
2-5. Using the relationship shown on Fig. 2-6 for heat transfer when a fluid flows inside tube, what is the percentage 
increase or decrease in the convection heat-transfer coefficient h
c
 if the viscosity of the fluid is decreased 10 percent. 
 
CHAPTER 2 - THERMAL PRINCIPLES 
Page 4 of 5 
Solution: 
 Figure 2-6. 
 
0.40.8Pr0.023ReNu = 
 where: 
 
µ
ρ
=
VD
Re
 
 k
c
Pr
pµ
=
 
 k
Dh
N c=u
 
 Then, 
 
0.4
p2
0.8
2
0.4
p1
0.8
1
c2
c1
k
cVD
0.023
k
cVD
0.023
k
Dh
k
Dh





µ






µ
ρ





 µ






µ
ρ
=












 
 
0.4
1
2
c2
c1
h
h






µ
µ
=
 
 If viscosity is decreased by 10 % 
 
0.9
1
2
=
µ
µ
 
 Then, 
 
( )0.4
c2
c1
h
h 9.0=
 
 h
c2
 = 1.043h
c1
 
 
 
( )100%
h
hh
Increase
c1
c1c2 −
=
 
 Increase = (1.043 - 1)(100 %) 
 Increase = 4.3 % - - - Ans. 
 
2-6. What is the order of magnitude of heat release by convection from a human body when the air velocity is 0.25 m/s and 
its temperature is 24 C? 
 
Solution: 
 Using Eq. (2-12) and Eq. (2-18) 
 
 C = h
c
A( t
s
 - t
a
 ) 
 h
c
 = 13.5V
0.6
 
 V = 0.25 m/s 
 h
c
 = 13.5(0.25)
0.6
 = 5.8762 W/m
2
.K 
 
CHAPTER 2 - THERMAL PRINCIPLES 
Page 5 of 5 
 Human Body: A = 1.5 to 2.5 m
2
 use 1.5 m
2
 
 t
s
 = 31 to 33 C use 31 C 
 
 C = (5.8762 W/m
2
.K)(1.5 m
2
)(31 C - 24 C) 
 C = 61.7 W 
 
 Order of Magnitude ~ 60 W - - - Ans. 
2-7 What is the order of magnitude of radiant heat transfer from a human body in a comfort air-conditioning situation? 
 
Solution: 
 
 Eq. 2-10. 
 
( )4241A21 TTFAFq −εσ=− 
 
 Surface area of human body = 1.5 to 2.5 m
2
 use 1.5 m
2
 
 AF
ε
F
A
 = (1.0)(0.70)(1.5 m
2
) - 1.05 m
2
 
 s = 5.669x10
-8
 W/m
2
.K
4
 
 T
1
=31 C + 273 = 304 K 
 T
2
 = 24 C + 273 = 297 K 
 
 q
1-2
 = (5.669x10
-8
)(1.05)(304
4
 - 297
4
) 
 q
1-2
 = 45 W 
 
 Order of Magnitude ~ 40 W - - - Ans. 
 
2-8. What is the approximate rate of heat loss due to insensible evaporation if the skin temperature is 32 C, the vapor 
pressure is 4750 Pa, and the vapor pressure of air is 1700 Pa? The latent heat of water is 2.43 MJ/kg; C
diff
 = 1.2x10
-9
 
kg/Pa.s.m
2
. 
 
Solution: 
 Equation 2-19. 
 q
ins
 = h
fg
AC
dif
f( p
s
 - p
a
 ) 
 
 Where: 
 A = 2.0 m
2
 average for human body area 
 h
fg
 = 2.43 MJ/kg = 2,430,000 J/kg 
 p
s
 = 4750 Pa 
 p
a
 = 1700 Pa 
 C
diff
 = 1.2x10
-9
 kg/Pa.s.m
2
 
 
 q
ins
 = (2,430,000)(2.0)(1.2x10
-9
)(4750 - 1700) 
 
 q
ins
 = 18 W - - - Ans. 
- 0 0 0 - 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 1 of 9 
3-1 Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air when the 
following conditions prevail: t = 30 C, W = 0.015 kg/kg, and p
t
 = 90 kPa. 
 
Solution: 
 Equation 3-4. 
 st
a
a
a
pp
TR
p
TR
−
==ν
 
 T = 30 C + 273 = 303 K 
 R
a
 = 287 J/kg.K 
 P
t
 = 90 kPa = 90,000 Pa 
 
 Equation 3-2 
 
 st
s
pp
0.622p
W
−
=
 
 s
s
p90
0.622p
0.015
−
=
 
 1.35 - 0.15p
s
 = 0.622p
s
 
 p
s
 = 2.1193 kPa 
 
 
( )( )
2119.390000
303287
pp
TR
st
a
−
=
−
=ν
 
 νννν = 0.99 m
3
/kg - - - Ans. 
 
3-2. A sample of air has a dry-bulb temperature of 30 C and a wet-bulb temperature of 25 C. The barometric 
pressure is 101 kPa. Using steam tables and Eqs. (3-2), (303), and (3-5), calculate (a) the humidity ration if 
this air is adiabatically saturated, (b) the enthalpy of air if it is adiabatically saturated, (c) the humidity ratio of 
the sample using Eq. (3-5), (d) the partial pressure of water vapor in the sample, and (e) the relative 
humidity. 
 
Solution: 
 Eq. 3-2. 
 st
s
pp
0.622pW
−
=
 
 Eq. 3-3. 
 h = c
p
t + Wh
g
 
 
 Eq. 3-5 
 h
1
 = h
2
 - (W
2
 - W
1
)h
f
 
 
 h
1
 = c
p
t
1
 + Wh
g1 
 
 
h
g1
 at 30 C = 2556.4 kJ/kg 
 t1 = 30 C 
 cp = 1.0 kJ/kg.K 
 
 h
1
 = (1)(30) + 2556.4W
1
 
 h
1
 = 30 + 2556.4W
1
 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 3 of 9 
 
 (e) At 30 C, p
s
 = 4.241 kPa 
 
 Relative Humidity = (2.84 kPa / 4.241 kPa)(100%) 
 
 Relative Humidity = 67 % - - - Ans. 
 
3-3 Using humidity ratios from the psychrometric chart, calculate the error in considering the wet-bulb line to be 
the line of constant enthalpy at the point of 35 C dry-bulb temperature and 50 percent relative humidity. 
 
Solution: 
 
 Dry-bulb Temperature = 35 C 
 Relative Humidity = 50 % 
 
 
 Fig. 3-1, Psychrometric Chart. 
 At constant enthalpy line: Wet-bulb = 26.04 C 
 At wet-bulb line = Wet-bulb = 26.17 C 
 
 Error = 26.17 C - 26.04 C 
 Error = 0.13 C 
 
 
3-4. An air-vapor mixture has a dry-bulb temperature pf 30 C and a humidity ratio of 0.015. Calculate at two 
different barometric pressures, 85 and 101 kPa, (a) the enthalpy and (b) the dew-point temperature. 
 
Solution: 
 
 At 30 C, p
s
 = 4.241 kPa, h
g
 = 2556.4 kJ/kg 
 
 (a) h = c
p
t + Wh
g
 
 
 For 85 and 101 kPa 
 c
p
 = 1.0 
 t = 30 C 
 W = 0.015 kg/kg 
 h
g
 = 2556.4 kJ/kg 
 
 h = (1.0)(30) + (0.015)(2556.4) 
 h = 68.3 kJ/kg 
 
 (b) For dew-point: 
 st
s
pp
0.622p
W
−
=
 
 
 at 85 kPa 
 
 st
s
pp
0.622p
0.015
−
=
 
 p
s
 = 2.0016 kPa 
 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 4 of 9 
 Dew-Point = 17.5 C - - - Ans. 
 
 at 101 kPa 
 
 st
s
pp
0.622p
0.015
−
=
 
 p
s
 = 2.3783 kPa 
 
 Dew-Point = 20.3 C - - - Ans. 
 
3-5. A cooling tower is a device that cools a spray of water by passing it through a stream of air. If 15 m
3
/s of air is 
at 35 C dry-bulb and 24 C wet-bulb temperature and an atmospheric pressure of 101 kPa enters the tower 
and the air leaves saturated at 31 C, (a) to what temperature can this airstream cool a spray of water 
entering at 38 C with a flow rate of 20 kg/s and (b) how many kilograms per second of make-up water must 
be added to compensate for the water that is evaporated? 
 
Solution: 
 At 35 C dry-bulb, 24 C wet-bulb. 
 
 Fig. 3-1, Psychrometric Chart 
 h
1
 = 71.524 kJ/kg, 
 ν
1
 = 0.89274 m
3
/kg 
 W
1
 = 0.0143 kg/kg 
 At 31 C saturated, Table A-2. 
 h
2
 = 105 kJ/kg 
 W
2
 = 0.0290 kg/kg 
 
 Then; 
 m = (15 m
3
/s) / (0.89274 m
3
/kg) = 16.8022 kg/s 
 
 (a) t
w1
 = 38 C 
 m
w
 = 20 kg/s 
 c
pw
 = 4.19 kJ/kg.K 
 
 m
w
c
pw
(t
w1
 - t
w2
) = m(h
2
 - h
1
) 
 
 
 (20)(4.19)(38 - t
w2
) = (16.8022)(105 - 71.524) 
 
 t
w2
 = 31.3 C - - - Ans. 
 
 (b) Make-Up Water = m
m
 
 
 m
m
 = m(W
2
 - W
1
) 
 m
m
 = (16.8022)(0.0290 - 0.0143) 
 
 m
m
 = 0.247 kg/s - - - Ans. 
 
 
3-6. In an air-conditioning unit 3.5 m
3
/s of air at 27 C dry-bulb temperature, 50 percent relative humidity, and 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 5 of 9 
standard atmospheric pressure enters the unit. The leaving condition of the air is 13 C dry-bulb temperature 
and 90 percent relative humdity. Using properties from the psychrometric chart, (a) calculate the refrigerating 
capacity inkilowatts and (b) determine the rate of water removal from the air. 
 
Solution: 
 
 At 27 C dry-buld, 5 Percent Relative Humidity 
 h
1
 = 55.311 kJ/kg, 
 ν
1
 = 0.86527 m
3
/kg 
 W
1
 = 0.0112 kg/kg 
 
 At 13 C Dry-Bulb, 90 Percent Relative Humidity 
 h
2
 = 33.956 kJ/kg 
 W
2
 = 0.0084 kg/kg 
 
 
 m = (3.5 m
3
/s)/(0.86526 m
3
/kg) = 4.04498 kg/s 
 
 (a) Refrigerating Capacity 
 = m(h
1
 - h
2
) 
 = (4.04498)(55.311 - 33.956) 
 = 86.38 kW - - - Ans. 
 
 (b) Rate of Water Removal 
 = m(W
1
 - W
2
) 
 = (4.04498)(0.0112 - 0.0084) 
 = 0.0113 kg/s - - - Ans. 
 
 
3-7. A stream of outdoor air is mixed with a stream of return air in an air-conditioning system that operates at 101 
kPa pressure. The flow rate of outdoor air is 2 kg/s, and its condition is 35 C dry-bulb temperature and 25 C 
wet-bulb temperature. The flow rate of return air is 3 kg/s, and its condition is 24 C and 50 percent relative 
humidity. Determine (a) the enthalpy of the mixture, (b) the humidity ratio of the mixture, (c) the dry-bulb 
temperature of the mixture from the properties determined in parts (a) and (b) and (d) the dry-bulb 
temperature by weighted average of the dry-bulb temperatures of the entering streams. 
 
Solutions: 
 
 Use Fig. 3-1, Psychrometric Chart 
 At 35 C Dry-Bulb, 24 C Wet-Bulb 
 h
1
 = 75.666 kJ/kg, m
1
 = 2 kg/s 
 W
1
 = 0.0159 kg/kg 
 
 At 24 C Dry-Bulb, 50 Percent Relative Humidity 
 h
2
 = 47.518 kJ/kg, m
2
 = 3 kg/s 
 W
2
 = 0.0093 kg/kg 
 
 (a) 
 
 
( )( ) ( )( )
32
47.518375.6662
hm +
+
=
 
 h
m
 = 58.777 kJ/kg - - - Ans. 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 6 of 9 
 
 (b) 
 
 
 
( )( ) ( )( )
32
0.009330.01592
Wm +
+
=
 
 W
m
 = 0.1194 kg/kg - - - Ans. 
 
 (c) At 58.777 kJ/kg and 0.01194 kg/kg. 
 From Psychrometric Chart, Fig. 3-1. 
 
 Dry-Bulb Temperature = 28.6 C - - - Ans. 
 
 (d) 
 
 
( )( ) ( )( )
32
243352
tm +
+
=
 
 t
m
 = 28.4 C - - - Ans. 
 
3-8. The air conditions at the intake of an air compressor are 28 C, 50 percent relative humidity, and 101 kPa. 
The air is compressed to 400 kPa, then sent to an intercooler. If condensation of water vapor from the 
compressed air is to be prevented, what is the minimum temperature to which the air can be cooled in the 
intercooler? 
 
Solution: At 28 C, p
s
 = 3.778 kPa 
 At 50 percent relative humidity, p
s
 = (0.5)(3.778 kPa) = 1.889 kPa 
 
 st
s
pp
0.622p
W
−
=
 
 
 Moisture ratio is constant 
 
 at 101 kPa 
 
 
( )
1.889101
1.8890.622
W
−
=
 
 
 W = 0.011855 kg/kg 
 
 at 400 kPa, determine p
s
 
 
 s
s
p
0.622p
0.011855
−
=
400
 
 
 p
s
 = 7.4812 kPa 
 
 From Table A-1. 
 
 Dew-Point = 40.3 C - - - Ans. 
 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 7 of 9 
3-9. A winter air-conditioning system adds for humidification 0.0025 kg/s of saturated steam at 101 kPa pressure 
to an airflow of 0.36 kg/s. The air is initially at a temperature of 15 C with a relative humidity of 20 percent. 
What are the dry- and wet-bulb temperatures of the air leaving the humidifier? 
 
Solution: 
 
 At 15 C Dry-Bulb, 20 Percent Relative Humidity 
 h
1
 = 20.021 kJ/kg 
 W
1
 = 0.0021 kg/kg 
 
 At 101 kPa steam, h
fg
 = 2675.85 kJ/kg 
 m
s
 = 0.0025 kg/s 
 m = 0.36 kg/s 
 m
s
 = m(W
2
 - W
1
) 
 0.0025 = 0.36(W
2
 - 0.002) 
 W
2
 = 0.00894 kg/kg 
 
 m(h
2
 - h
1
) = m
s
h
g
 
 (0.36)(h2 - 20.021) = (0.0025)(2675.85) 
 h
2
 = 38.6 kJ/kg 
 
 Fig. 3-1, Psychrometric Chart 
 W
2
 = 0.00894 kg/kg 
 h
2
 = 38.6 kJ/kg 
 
 Dru-Bulb Temperature = 16.25 C 
 Wet-Bulb Temperature = 13.89 C 
 
3-10. Determine for the three cases listed below the magnitude in watts and the direction of transfer of sensible 
heat [ using Eq. (3-8)], latent heat [ using Eq. (3-9)], and total heat [ using Eq. (3-14)]. the area is 0.15 m
2
 and 
h
c
 = 30 W/m
2
.K. Air at 30 C and 50 percent relative humidity is in contact with water thatis at a temperature 
of (a) 13 C, (b) 20 C, and (c) 28 C. 
 
Solution: 
 Equation 3-8. 
 dq
s
 = h
c
dA(t
i
 - t
a
) 
 Equation 3-9. 
 dq
L
 = h
D
dA(W
i
 - W
a
)h
fg 
 
Equarion 3-14.
 
 
)h(h
c
dAh
dq ai
pm
c
t −=
 
 At 30 C, 50% Relative Humidity 
 h
a
 = 63.965 kJ/kg = 63,965 J/kg 
 W
a
 = 0.0134 kg/kg 
 
 (a) 13 C 
 
 dq
s
 = h
c
dA(t
i
 - t
a
) 
 dq
s
 = (30)(0.15)(13 - 30) 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 8 of 9 
 dq
s
 = -76.5 W - - - Ans. 
 
 dq
L
 = h
D
dA(W
i
 - W
a
)h
fg 
 W
i
 at 13 C = 0.00937 kg/kg from Table A-2 
 h
fg
 at 13 C = 2,470,840 J/kg 
 h
D
 = h
c
 / c
pm
 
 c
pm
 = 1020 kJ/kg.K 
 h
D
 = 30 / 1020 = 0.029412
 
 
 
dq
L
 = (0.029412)(0.15)(0.00937 - 0.0134)(2,470,840) 
 dq
L
 = -43.93 W - - - Ans. 
 h
i
 at 13 C = 36,719 J/kg from Table A-2 
 
 
( )( ) ( )63,96536,719
1020
0.1530
)h(h
c
dAh
dq ai
pm
c
t −=−=
 
 
 dq
t
 = -120.2 W - - - Ans. 
 
 (b) 20 C 
 
 dq
s
 = h
c
dA(t
i
 - t
a
) 
 dq
s
 = (30)(0.15)(20 - 30) 
 dq
s
 = -45 W - - - Ans. 
 
 dq
L
 = h
D
dA(W
i
 - W
a
)h
fg 
 W
i
 at 20 C = 0.01475 kg/kg from Table A-2 
 h
fg
 at 20 C = 2,454,340 J/kg 
 h
D
 = h
c
 / c
pm
 
 c
pm
 = 1020 kJ/kg.K 
 h
D
 = 30 / 1020 = 0.029412
 
 
 
dq
L
 = (0.029412)(0.15)(0.01475 - 0.0134)(2,454,340) 
 dq
L
 = 14.62 W - - - Ans. 
 
 h
i
 at 20 C = 57,544 J/kg from Table A-2 
 
 
( )( ) ( )63,96557,544
1020
0.1530
)h(h
c
dAh
dq ai
pm
c
t −=−=
 
 
 dq
t
 = -28.33 W - - - Ans. 
 
 
 (c) 28 C 
 
 dq
s
 = h
c
dA(t
i
 - t
a
) 
 dq
s
 = (30)(0.15)(28 - 30) 
 dq
s
 = -9.0 W - - - Ans. 
CHAPTER 3- PSYCHROMETRY AND WETTED-SURFACE HEAT TRANSFER 
Page 9 of 9 
 
 dq
L
 = h
D
dA(W
i
 - W
a
)h
fg 
 W
i
 at 28 C = 0.02422 kg/kg from Table A-2 
 h
fg
 at 28 C = 2,435,390 J/kg 
 h
D
 = h
c
 / c
pm
 
 c
pm
 = 1020 kJ/kg.K 
 h
D
 = 30 / 1020 = 0.029412
 
 
 
dq
L
 = (0.029412)(0.15)(0.02422 - 0.0134)(2,435,390) 
 dq
L
 = 116.3 W - - - Ans. 
 
 h
i
 at 28 C = 89,952 J/kg from Table A-2 
 
 
( )( ) ( ),96589,952
1020
0.1530
)h(h
c
dAh
dq ai
pm
c
t 63−=−=
 
 
 
 dq
t
 = 114.6 W - - - Ans. 
- 0 0 0 - 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 1 of 8 
4-1 The exterior wall of a single-story office building near Chicago is 3 m high and 15 m long. The wall consists 
of 100-mm facebrick, 40-mm polystyrene insulating board. 150-mm lightweight concrete block, and an 
interior 16-mm gypsum board. The wall contains three single-glass windows 1.5 n high by 2 m long. 
Calculate the heat loss through the wall at design conditions if the inside temperature is 20 C. 
 
Solution: 
 
 Table 4-3, Design Outdoor is -18 C for Chicago. 
 
 
 For the wall: 
 Area, A = (3 m)(15 m) - (3)(1.5 m)(2 m) = 36 m
2
. 
 
 Resistance: Table 4-4. 
 
 Outside Air Film 0.029 
 Facebrick, 100 mm 0.076 
 Polystyrene Insulating Board, 40 mm 1.108 
 Lightweight Concrete Block, 150 mm 0.291 
 Gypsum Board, 16 mm 0.100 
 Inside Air Film 0.120 
 ==== 
 R
tot
 = 1.724 m
2
.K/ W 
 
 Wall: 
 
( )2018
1.724
36
t
Rtot
A
q −−=∆=
 
 q = -794 Watts 
 
 For the glass: 
 Area A = (3)(1.5 m0(2 m) = 9 m
2
 
 Table 4-4, U = 6.2 W/m
2
.K 
 q = UA∆t = (6.2)(9)(-18 - 20) 
 q = -2,120 Watts 
 
 Total Heat Loss Thru the Wall = -794 W -2,120 W 
 = -2,194 Watts - - - Ans 
. 
4-2. For the wall and conditions stated in Prob. 4-1 determine the percent reduction in heat loss through the wall if 
(a) the 40 mm of polystyrene insulation is replaced with 55 mm of cellular polyurethane, (b) the single-glazed 
windows are replaced with double-glazed windows with a 6-mm air space. (c) If you were to choose between 
modification (a) or (b) to upgrade the thermal resistance of the wall, which would you choose and why? 
 
Solution 
(a) Resistance: Table 4-4 
 Outside Air Film 0.029 
 Facebrick, 100 mm 0.076 
 Cellular Polyurethane, 55 mm 2.409 
 Lightweight Concrete Block, 150 mm 0.291 
 Gypsum Board, 16 mm 0.100 
 Inside Air Film 0.120 
 ===== 
 R
tot
 = 3.025 m
2
.K/W 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 2 of 8 
 Wall: 
 
( )2018
3.025
36
t
Rtot
A
q −−=∆=
 
 q = - 452 Watts 
 
 New Total Heat Loss Thru Wall 
 q = - 452 W - 2,120 W 
 q = - 2,572 W 
 
( ) ( ) ( )%100
W2,914
W2,572W2,914
%Reduction
−
−−−
=
 
 % Reduction = 11.74 %- - - Ans. 
 
 (b) For the glass: (Double-Glazed) 
 Table 4-4, U = 3.3 W/m
2
.K 
 
 q = UA∆t = (3.3)(9)(-18 - 20) 
 q = -1,129 Watts 
 
 New Total Heat Loss Thru Wall 
 q = - 794 W - 1,129 W 
 
( ) ( ) ( )%100
W2,914
W,923W2,914
%Reduction
−
−−−
=
1
 
 % Reduction = 34 %- - - Ans. 
 
 
(c) Choose letter b --- Ans. 
 
 
4-3 An office in Houston, Texas, is maintained at 25 C and 55 percent relative humidity. The average occupancy 
is five people, and there will be some smoking. Calculate the cooling load imposed by ventilation 
requirements at summer design conditions with supply air conditions set at 15 C and 95 percent relative 
humidity if (a) the recommended rate of outside ventilation air is used and (b) if a filtration device of E = 70 
precent is used. 
 
Solution: 
 Table 4-3, Houston Texas 
 Summer Deisgn Conditions 
 Dry-Bulb = 34 C 
 Wet-Bulb = 25 C 
 At 34 C Dry-Bulb, 24 C Wet-Bulb 
 h
o
 = 76 kJ/kg. W
o
 = 0.0163 kg/kg 
 At 15 C Dry-Bulb, 95 percent relative humidity 
 h
s
 = 40.5 kJ/kg, W
s
 = 0.010 kg/kg 
 At 25 C, 55 percent relative humidity 
 h
i
 = 53.2 kJ/kg, W
i
 = 0.011 kg/kg 
 
 (a) V = Vo 
 Table 4-1, 10 L/s per person 
 V = (10 L/s)(5) = 50 L/s 
 q
s
 = 1.23V(t
o
 - t
s
) 
 q
s
 = 1.23(50)(34- 15) 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 3 of 8 
 q
s
 = 1,168.5 W 
 
 q
L
 = 3000V(W
o
 - W
i
) 
 q
L
 = 3000(50)(0.0163 - 0.010) 
 q
L
 = 945 W 
 
 
 q
t
 = q
s
 + q
L 
 q
t
 = 1,168.5 W + 945 W 
 q
t
 = 2,113.5 W 
 q
t
 = 2.1 kw - - - Ans. 
 
 (a) V1 = Vm 
 Table 4-1, 2.5 L/s per person 
 V1 = (2.5 L/s)(5) = 12.5 L/s 
 E
VV
VV mor2
−
==
 
 0.7
12.550
V2
−
=
 
 V
2
 = 53.5714 L/s 
 
 q
s
 = 1.23V
1
(t
o
 - t
s
) + 1.23V
2
(t
i
 - t
s
) 
 q
s
 = 1.23(12.5)(34 - 15) + 1.23(53.5714)(25 - 15) 
 q
s
 = 951 W 
 
 q
L
 = 3000V
1
(W
o
 - W
s
) + 3000V
2
(W
i
 - W
s
) 
 q
L
 = 3000(12.5)(0.0163 - 0.010) + 3000(53.5714)(0.011 - 0.010) 
 q
L
 = 397 W 
 
 q
t
 = q
s
 + q
L 
 q
t
 = 951 W + 397 W 
 q
t
 = 1,348 W 
 q
t
 = 1.35 kw - - - Ans. 
 
 
4-4 A computer room located on the second floor of a five-story office building is 10 by 7 m. The exterior wall is 
3.5 m high and 10 m long; it is a metal curtain wall (steel backed with 10 mm of insulating board), 75 mm of 
glass-fiber insulation, and 16 mm of gypsum board. Single-glazed windows make up 30 percent of the 
exterior wall. The computer and lights in the room operate 24 h/d and have a combined heat release to the 
space of 2 kw. The indoor temperature is 20 C. 
 (a) If the building is located in Columbus, Ohio, determine the heating load at winter design conditions. 
 (b) What would be the load if the windows were double-glazed? 
 
Solution:(a) Table 4-3, Columbus, Ohio, Winter Design Temperature = -15 C. 
 
 Thermal Transmission: 
 Wall: 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 4 of 8 
 
( )io t - t
Rtot
A
q =
 
 A = (3.5 m)(10 m)(0.70) = 24.5 m
2
 
 
 Table 4-4: 
 Outside Air Film 0.029 
 Insulating Board, 10 mm 0.320 
 Glass-Fiber Insulation, 75 mm 2.0775 
 Gypsum Board, 16 mm 0.100 
 Inside Air Film 0.120 
 ==== 
 R
tot
 = 2.6465 m
2
.K/W 
 
 
( )20- 15-
2.6465
24.5
qw =
 
 q
w
 = -324 W 
 
 
 Glass: 
 
 q = UA(to - ti) 
 A = (3.5 m)(10 m)(0.30) = 10.5 m2 
 Table 4-4. 
 Single Glass, U = 6.2 W/m
2
.K 
 q
G
 = (6.2)(10.5)(-15 - 20) 
 q
G
 = -2,278.5 W 
 
 q
t
= -324 W - 2,278.5 W = -2,602.5 W 
 
 Heating Load = 2,602.5 W - 2,000 W 
 Heating Load = 602.5 W - - - Ans. 
 
 
 (b) If double-glazed, Say 6-mm air space 
 Table 4-4, U = 3.3 W/m
2
.K 
 
 qG = (3.3)(10.5)(-15 - 20) 
 qG = -1,212.8 W 
 
 qt = -324 W - 1,212.8 W = -1,536.8 W 
 
 Since 1,536.8 W < 2,000 W, there is no additional heat load required. 
 
4-5. Compute the heat gain for a window facing southeast at 32
o
 north latitude at 10 A.M. central daylight time on 
August 21. The window is regular double glass with a 13-mm air space. The glass and inside draperies have 
a combined shading coefficient of 0.45. The indoor design temperature is 25 C, and the outdoor temperature 
is 37 C. Window dimensions are 2 m wide and 1.5 m high. 
 
Solution: 
 Window Area = 2 m x 1.5 m = 3.0 m
2
 
 Table 4-4, U = 3.5 W/m
2
.K 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 5 of 8 
 
 
 Transmission: 
 q
1
 = UA(t
o
 -t
i
) 
 q
1
 = (3.5)(3)(37 - 25) 
 q
1
 = 126 W 
 
 
 Solar: 
 q
sg
 = (SHGFmax)(SC)(CLF)A 
 
 Table 4-10, 32
o
 North Latitude, Facing SE 
 SHGF = 580 W/m
2
 
 Table 4-12, Facing SE at 10 A.M. 
 CLF = 0.79 
 
 and SC = 0.45 
 
 q
sg
 = (580)(0.45)(0.79)(3) 
 q
sg
 = 618.6 W 
 
 Heat Gain = 126 W + 618.6 W 
 Heat Gain = 744.6 W - - - Ans. 
 
4-6. The window in Prob. 4-5 has an 0.5-m overhang at the top of the wiindow. How far will the shadow extend 
downward? 
 
Solution: 
 From Fig. 4-5 
 
γ
β
=
cos
tan
dy
 
 d = 0.5 m 
 Table 4-3, 32
o
 North Latitude, 10 A.M., August 
 β = 56o 
 φ = 60o 
 
 
 Facing South East, ψ = 45o 
 γ = φ - ψ = 60 - 45 = 45o 
 
 
( )
o
o
cos15
tan56
m0.5
cos
tan
dy =
γ
β
=
 
 y = 0.77 m - - - Ans. 
 
4-7. Compute the instantaneous heat gain for the window in Prob. 4-5 with the external shade in Prob. 4-6. 
 
Solution: 
 
 A
1
 = Sunlit Area = (2.0 m)(1.5 m - 0.77 m) = 1.46 m
2
 
 A = 3.0 m
2
 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 6 of 8 
 
 Transmission = UA(to - t1) 
 = (3.5)(3)(37 - 25) 
 = 126 W 
 
 Solar: 
 q
sg
 = (SHGF
max
)(SC)(CLF)A
1
 
 q
sg
 = (580)(0.45)(0.79)(1.46) = 301 W 
 
 Heat Gain = 126 W + 301 W = 427 W - - - Ans. 
 
4-8. Compute the total heat gain for the south windows of an office building that has no external shading. The 
windows are double-glazed with a 6-mm air space and with regular plate glass inside and out. Draperies with 
a shading coefficient of 0.7 are fully closed. Make Calculation for 12 noon in (a) August and (b) December at 
32
o
 North Latitude. The total window area is 40 m
2
. Assume that the indoor temperatures are 25 and 20 C 
and that the outdoor temperatures are 37 and 4 C. 
 
Solution: 
 Tabkle 4-7 
 Double-glazed, 6-mm air space, U-value 
 Summer - 3.5 W/m
2
.K 
 Winter - 3.3 W/m
2
.K 
 
 A = 40 m
2
 
 
 (a) August, SUmmer, Indoor = 25 C, Outdoor = 37 C 
 
 Thermal Transmission: 
 q
1
 = UA(t
o
 -t
i
) 
 q
1
 = (3.5)(40)(37 - 25) 
 
 q
1
 = 1,680 W 
 
 
 Solar: 
 q
sg
 = (SHGFmax)(SC)(CLF)A 
 
 Table 4-10, 32
o
 North Latitude, Facing South 
 SHGF = 355 W/m
2
 
 Table 4-12, Facing South at 12 Noon. 
 CLF = 0.83 
 
 and SC = 0.7 
 
 q
sg
 = (355)(0.7)(0.83)(40) 
 q
sg
 = 8,250 W 
 
 q
t
 = q
1
 + q
sg
 
 q
t
 = 1,680 W + 8,250 W 
 q
t
 = 9,930 W - - - Ans. 
 
 (b) December, Winter, Indoor = 20 C, Outdoor = 4 C 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 7 of 8 
 
 Thermal Transmission: 
 q
1
 = UA(t
o
 -t
i
) 
 q
1
 = (3.3)(40)(4 - 20) 
 
 q
1
 =-2,112 W 
 
 
 Solar: 
 q
sg
 = (SHGFmax)(SC)(CLF)A 
 
 Table 4-10, 32
o
 North Latitude, Facing South, December 
 SHGF = 795 W/m
2
 
 Table 4-12, Facing South at 12 Noon. 
 CLF = 0.83 
 
 
 and SC = 0.7 
 
 q
sg
 = (795)(0.7)(0.83)(40) 
 q
sg
 = 18,476 W 
 
 q
t
 = q
1
 + q
sg
 
 q
t
 = -2,112 W + 18,476 W 
 q
t
 = 16,364 W - - - Ans. 
 
 
4-9. Compute the instantaneous heat gain for the south wall of a building at 32
o
 north latitude on July 21. The 
time is 4 p.m. sun time. The wall is brick veneer and frame with an overall heat-transfer coefficient of 0.35 
W/m
2
.K. The wall is 2.5 by 6 m with a 1- x 2-m window. 
 
Solution: 
 
 Wall: A = (2.5 m)(5 m ) - (1 m)(2 m) = 10.5 m
2
 
 U = 0.35 W/m
2
.K 
 
 q
w
 = UA(CLTD) 
 Table 4-11, South, Type F, 4 P.M. 
 CLTD = 22 
 q
w
 = (0.35)(10.5)(22) 
 q
w
 = 80.85 Watts. - - - Ans 
 
4-10. Compute the peak instantaneous heat gain per square meter of area for a brick west wall similar to that in 
Example 4-3. Assume that the wall is located at 40
o
 north latitude. The date is July. What time of the day 
does the peak occur? The outdoor daily average temperature of 30 C and indoor design temperature is 25 C. 
 
Solution: 
 Ex. 4-3, U = 0.356 W/m
2
.K 
 Table 4-15, Type F, West Wall 
 CLTD
max
 = 33 at 1900 h or 7 P.M. 
CHAPTER 4 - HEATING AND COOLING LOAD CALCULATION 
Page 8 of 8 
 
 CLTD
adj
 = CLTD + (25 - T
i
) + (T
av
 - 29) 
 CLTD
adj
 = 33 + (30 - 29) = 34 C 
 
 q
max
 / A = U(CLTD) 
 q
max
 / A = (0.356)(34) 
 q
max
 / A = 12.1 W/m
2
 at 7 P.M. - - - - Ans. 
 
- 0 0 0 - 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 1 of 4 
5-1 A conditioned space that is maintained at 25 C and 50 percent relative humidity experience a sensible-heat 
of 80 kW and a latent-heat gain of 34 kW. At what temperature does the load-ratio line intersect the 
saturation line? 
 
Solution: 
 
 LS
S
qq
q
ratioLoad
+
=−
 
 q
S
 = 80 kW 
 q
L
 = 34 kW 
 
 3480 +
=−
80
ratioLoad
 
 
 Load-ratio = 0.7018 
 
 But, 
 
 
( )
ratioLoad
h-h
t-tc
ic
icp
−=
 
 
 At 25 C, 50 percent relative humidity 
 h
c
 = 50.5 kJ/kg 
 
 Try t
i
 = 15 C 
 
 
( )
0.7018
h-h
t-tc
ic
icp
=
 
 
 
( )( )
0.7018
h-50.5
15-251.0
i
=
 
 
 Connecting the two-points gives the load-ratio line which intersects the saturation line at 9.75 C with hi = 
28.76 kJ/kg. 
 
 Ans. 9.75 C. 
 
 
5-2. A conditioned space receives warm, humidified air during winter air conditioning in order to maintain 20 C 
and 30 percent relative humidity. The space experiences an infiltration rate of 0.3 kg/s of outdoor air and an 
additional sensible-heat loss of 25 kW. The outdoor air is saturated at a temperature of -20 C (see Table A-
2). If conditioned air is supplied at 40 C dry-buld, what must be the wet-bulb temperature of supply air be in 
order to maintain the space conditions? 
 
Solution: 
 
 At -20 saturated, 
 h
1
 = -18.546 kJ/kg 
 m
1
 = 0.3 kg/s 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 2 of 4 
 
 Additional heat loss = 25 kW 
 
 At 20 C and 30 percentrelative humidity, 
 h
2
 = 31 kJ/kg 
 
 t
3
 = 40 C 
 
 Equations: 
 
 Sensible Heat Balance: 
 
 m
2
(t
3
 - t
2
) + m
1
(t
1
 - t
2
) = 25 kW 
 m
2
(40 - 20) + (0.3)(-20 - 20) = 25 
 m
2
 = 1.85 kg/s 
 
 Total Heat Balance: 
 m
2
(h
3
 - h
2
) + m
1
(h
1
 - h
2
) = 25 kW 
 (1.85)(h
3
 - 31) + (0.3)(-18.546 - 31) = 25 
 h
3
 = 52.55 kJ/kg 
 
 Then at 40 C and 52.55 kJ/kg. 
 
 Wet-Bulb Temperature = 18.8 C - - - Ans. 
 
5-3. A laboratory space to be maintained at 24 C and 50 percent relative humidity experiences a sensible-cooling 
load of 42 kW and a latent load of 18 kW. Because the latent load is heavy, the air-conditioning system is 
equipped for reheating the air leaving the cooling coil. The cooling coil has been selected to provide outlet air 
at 9.0 C and 95 percent relative humidity. What is (a) the temperature of supply air and (b) the airflow rate? 
 
Solution: 
 
 q
S
 = 42 kW 
 q
L
 = 18 kw 
 
 At 24 C , 50 percent relative humidity 
 h
i
 = 47.5 kJ/kg 
 
 At 9.0 C, 95 percent relative humidity 
 h
c
 = 26 kJ/kg 
 
 
( )
ic
icp
hh
ttc
lineratioloadCoil
−
−
=−
 
 
( )( ) 70.024 =−=−
47.5-26
91.0
lineratioloadCoil
 
 
0.70
1842
42
qq
q
lineratioloadCoil
LS
S
=
+
=
+
=−
 
 
 (a) Since 9 C < 13 C minimum. 
 Temperature of supply air = 13 C - - - Ans. 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 3 of 4 
 
 (b) 
 
( ) ( )( )13241.0
42
tt1.0
q
m
21
S
−
=
−
=
 
 m = 3.82 kg/s - - - - Ans. 
 
5-4. In discussing outdoor-air control Sec. 5-3 explained that with outdoor conditions in the X and Y regions on 
the psychrometric chart in Fig. 5-5 enthalpy control is more energy-efficient. We now explore some 
limitations of that statement with respect to the Y region. Suppose that the temperature setting of the outlet 
air form the cooling coil is 10 C and that the outlet air is essentially saturated when dehumidification occurs in 
the coil. If the condition of return air is 24 C and 40 percent relative humidity and the outddor conditions are 
26 C and 30 percent relative humidity, would return air or outside air be the preferred choice? Explain why. 
 
Solution: 
 
 See Fig. 5-5 and Sec. 5-3. 
 
 Outside Air: At 26 C, 30 percent relativw humidity 
 h
o
 = 42 kJ/kg 
 Coil outlet = 10 C saturated 
 q = 42 kJ/kg - 29.348 kJ/kg 
 q = 12.652 kJ/kg 
 
 Recirculated air: At 24 C, 40 percent relative humidity 
 h
i
 = 43 kJ/kg 
 With 10% outdoor air. 
 h
m
 = (0.10)(42) + (0.90)(43) = 42.9 kJ/kg 
 q = 42.9 kJ/kg - 29.348 kJ/kg 
 q = 13.552 kJ/kg > 12.652 kJ/kg. 
 
Ans. Outside air is preferred due to lower cooling required. 
 
 
5-5. A terminal reheat system (Fig. 5-9) has a flow rate of supply air of 18 kg/s and currently is operating with 3 
kg/s of outside air at 28 C and 30 percent relative humidity. The combined sensible load in the spaces is 140 
kw, and the latent load is negligible. The temperature of the supply air is constant at 13 C. An accountant of 
the firm occupying the building was shocked by the utility bill and ordered all space thermostat be set up from 
24 to 25 C. What is the rate of heat removal in the cooling coil before and after the change and (b) the rate of 
heat supplied at the reheat coils before and after change? Assume that the space sensible load remains at 
140 kw? 
 
Solution: See Fig. 5-9. 
 
 Outside air at 28 C and 30 percent relative humidity 
 h
o
 = 46 kJ/kg 
 
 At 24 C Set-Up. 
 Coil entering temperature, t
m
 
 t
m
 = [(3)(28) + (18 - 3)(24)] / 18 = 24.667 C 
 
 Coil supply temperature = 13 C constant 
 Cooling rate = (18)(24.667 - 13) = 210 kw 
 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 4 of 4 
 Space sensible load = 140 kw constant 
 
 Reheat supply temperature, t
s
. 
 t
s
 = 24 - 140 / 18 = 16.222 C 
 Heating Rate = (18)(16.222 - 13) 
 Heating Rate = 58 kw 
 
 At 25 C Set-Up. 
 Coil entering temperature, t
m
 
 t
m
 = [(3)(28) + (18 - 3)(25)] / 18 = 25.5 C 
 
 Coil supply temperature = 13 C constant 
 Cooling rate = (18)(25.5 - 13) = 225 kw 
 
 Space sensible load = 140 kw constant 
 
 Reheat supply temperature, t
s
. 
 t
s
 = 25 - 140 / 18 = 17.222 C 
 Heating Rate = (18)(17.222 - 13) 
 Heating Rate = 76 kw 
 
Answer: 
 
 (a) Before = 210 kw 
 After = 225 kw 
 15 kw increase in cooling rate. 
 (b) Before = 58 kw 
 After = 76 kw 
 18 kw increase in heating rate 
 
- 0 0 0 - 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 1 of 13 
6-1. Compute the pressure drop of 30 C air flowing with a mean velocity of 8 m/s in a circular sheet-metal duct 
300 mm in diameter and 15 m long using (a) Eqs. (6-1) and (6-2) and (b) Fig. 6-2. 
 
Solution: 
 Equation 6-1. 
 
ρ=∆
2
V
D
L
fp
2
 
 Equation 6-2. 
 
( )
2
f
D
Re
9.3D
2log14
1
f


























ε
+−
ε
+
=
1log2.1
 
 
 D = 300 mm = 0.3 m 
 V = 8 m/s 
 
 At 30 C, Table 6-2. 
 µ = 18.648 mPa.s = 1.8648 x10-5 Pa.s 
 ρ = 1.1644 kg/m3 
 
 Table 6-1, ε = 0.00015 m 
 
 ε/D = 0.00015 m / 0.3 m = 0.0005 
 
 Reynolds Number 
 
 
µ
ρ
=
VD
Re
 
 
( )( )( )
149,860
101.8648
1.16440.38
Re
5
=
×
=
−
 
 (a) Equation 6-2. 
 
( )( )
2
f0.0005149860
9.3
12log
0.0005
1
2log1.14
1
f




















+−





+
=
 
 
2
f
0.124116
12log7.74206
1
f




















+−
=
 
 
 By trial and error; f = 0.01935 
 
 Equation 6-1 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 2 of 13 
 
( ) ( ) ( )1.1644
2
8
0.3
15
0.1935p
2






=∆
 
 
 
 ∆∆∆∆p = 36 Pa - - - Ans. 
 
 (b) From Fig. 6-2, D = 0.30 m, V = 8 m /s 
 
 Friction Loss = 2.57 Pa/m 
 
 For 15 m 
 
 
 
 ∆∆∆∆p = (15)(2.57) = 38.55 Pa - - - Ans. 
 
 
6-2. A pressure difference of 350 Pa is available to force 20 C air through a circular sheet-metal duct 450 mm in 
diameter and 25 m long. Using Eq. (6-1) and Fig. 6-1, determine the velocity. 
 
Solution: 
 Eq. 6-1 
 
ρ=∆
2
V
D
L
fp
2
 
 D = 450 mm = 0.45 m 
 Table 6-1, ε = 0.00015 
 ε/D = 0.00015 / 0.45 = 0.00033 
 
 At 20 C, Table 6-2. 
 µ = 18.178 mPa.s = 1.8178 x 10-5 Pa.s 
 ρ = 1.2041 kg/m3 
 
( )( )( )
5101.8178
1.2041V0.45VD
Re
−×
=
µ
ρ
=
 
 Re = 29,808 V 
 
ρ=∆
2
V
D
L
fp
2
 
 
( )1.2041
2
V
0.45
25
f350
2






=
 
 
 fV
2
 = 10.46425 
 
 Use Eq. 6-2. 
 
 
( )
2
f
D
Re
9.3D
2log14
1
f


























ε
+−
ε
+
=
1log2.1
 
 f
3.23485
V =
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 3 of 13 
 
( )
f
96,424.4
f
3.23485
29,808Re =





=
 
 Then: 
 
( )
2
f0.00033
f
96,424.4
9.3
12log
0.00033
1
2log1.14
1
f












































+−





+
=
0.0161
0.2227066.962971.14
1
f
2
=





−+
=
 
 0.0161
3.23485
V =
 
 V = 25.5 m/s - - - Ans. 
 
6-3 A rectangular duct has dimensions of 0.25 by 1 m. Using Fig. 6-2, determinethe pressure drop per meter 
length when 1.2 m
3
/s of air flows through the duct. 
 
Solution: 
 Q = 1.2 m
3
/s 
 a = 0.25 m 
 b = 1 m 
 
 Using Fig. 6-2. 
 Eq. 6-8. 
 
( )
( )0.25
0.625
eq,f
ba
ab
1.30D
+
=
 
 
( )
( )0.25
0.625
eq,f
1.00.25
1.00.25
1.30D
+
×
=
 
 
m0.517De q,f = 
 
 Fig. 6-2: Q = 1.2 m
3
/s, 
 
m0.517De q,f = 
 
 Then ∆∆∆∆p = 0.65 Pa/m - - - Ans. 
 
6-4. A sudden enlargement in a circular duct measures 0.2 m diameter upstream and 0.4 m diameter 
downstream. The upstream pressure is 150 Pa and downstream is 200 Pa. What is the flow rate of 20 C air 
through the fitting? 
 
 
Solution: 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 4 of 13 
 Equation 6-11: 
 
Pa
A
A
1
2
V
p
2
2
1
2
1
loss 





−
ρ
=
 
 
( )
loss
2
2
2
1
12 p
2
VV
pp −
ρ−
=−
 
 Table 6-2. 
 At 20 C, ρ = 1.2041 kg/m3 
 
 A
1
V
1
 = A
2
V
2
 
 D
1
 = 0.2 m 
 D
2
 = 0.4 m 
 D
1
2
V
1
 = D
2
2
V
2
 
 (0.2)
2
V
1
 = (0.4)
2
V2 
 V
2
 = 0.25V
1 
 
0.25
D
D
A
A
2
2
2
1
2
1
==
 
 p
2
 - p
1
 = 200 Pa - 150 Pa = 50 Pa 
 
( )
Pa
A
A
1
2
V
2
VV
pp
2
2
1
2
1
2
2
2
1
12 





−
ρ
−
ρ−
=−
 
 
( )[ ]( ) ( ) ( )2212121 0.251
2
1.2041V
2
1.20410.25VV
50 −−
−
=
 
 V
1
 = 14.88171 m/s 
 
 Q = A1V1 
 1
2
14
VDQ pi=
 
 
( ) ( )14.881710.2Q 2
4
pi
=
 
 Q = 0.4675 m
3
/s - - - Ans. 
 
6-5. A duct 0.4 m high and 0.8 m wide, suspended from a ceiling in a corridor, makes a right-angle turn in 
horizontal plane. The inner radius is 0.2 m, and the outer radius is 1.0 m, measured from the same center. 
The velocity of air in the duct is 10 m/s. To how many meters of straight duct is the pressure loss in this 
elbow equivalent? 
 
Solution: 
 
 Inner radius = 0.2 m 
 Outer radius = 1.0 m 
 
 W = 0.8 m 
 H = 0.4 m 
 
 Figure 6-8: 
 W / H = 0.8 / 0.4 = 2.0 
 Ratio of inner to outer radius = 0.2 / 1.0 = 0.2 
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 5 of 13 
 Then: 
 
0.35
2
V
p
2
loss
=
ρ
 
 
( )( )
m0.533
0.40.8
0.40.82
ba
2ab
Deq =+
=
+
=
 
 
 Friction loss for the D
eq
 = 1.95 Pa/m 
 
 Then: 
 
 
ρ=
2
V
0.35p
2
loss
 
 
 ρ = 1.2041 kg/m3 
 
 
( ) ( ) Pa211.2041
2
10
0.35p
2
loss ==
 
 
 
 Equivalent Length = 21 Pa / (1.95 Pa/m) 
 
 Equivalent lengtn = 10.8 m - - - - Ans. 
 
6-6. An 0.3- by 0.4-m branch duct leaves an 0.3- by 0.6-m main duct at an angle of 60
o
. The air temperature is 20 
C. The dimensions of the main duct remain constant following the branch. The flow rate upstream is 2.7 m
3
/s, 
and the pressure is 250 Pa. The branch flow rate is 1.3 m
3
/s. What is the pressure (a) downstream in the 
main duct and (b) in the branch duct? 
 
Solution: 
 p1 = 250 Pa, See Fig. 6-10. 
 β = 60o 
 ( )( )
m/s15
0.60.3
2.7
Vu ==
 
 ( )( )
m/s7.78
0.60.3
1.3-2.7
Vd ==
 
 ( )( )
m/s10.83
0.40.3
1.3
Vb ==
 
 
 at 20 C, ρ = 1.2041 kg/m3. 
 
 
 (a) Eq. 6-16. 
 
 
( ) Pa
V
V
10.4
2
V
p
2
u
d
2
d
loss 





−
ρ
=
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 6 of 13 
 
( ) ( ) ( ) Pa
15
7.78
10.4
2
1.20417.78
p
22
loss 





−=
 
 
 p
loss
 = 3.377 Pa 
 
 Bernoulli Equation 6-10 
 








ρ
−−+
ρ
ρ= loss
2
2
2
11
2
p
2
V
2
Vp
p
 
 
( ) 





−−+=
1.2041
3.377
2
7.78
2
15
1.2041
250
1.2041p
22
2
 
 
 p
2
 = 346 Pa - - - Ans. 
 
 (b) Fig. 6-11 
 
0.722
15
10.83
V
V
u
b
==
 
 β = 60o 
 
583.1
2
V
p
2
loss
=
ρ
 
 
( ) ( ) Pa111.81.2041
2
10.83
1.583p
2
loss ==
 
 
 








ρ
−−+
ρ
ρ= loss
2
2
2
11
2
p
2
V
2
Vp
p
 
 
( ) 





−−+=
1.2041
111.8
2
10.83
2
15
1.2041
250
1.2041p
22
2
 
 
 p
2
 = 203 Pa - - - Ans. 
 
 
6-7. In a branch entry, an airflow rate of 0.8 m
3
/s joins the main stream to give a combined flow rate of 2.4 m
3
/s. The air 
temperature is 25 C. The branch enters with an angle of β = 30o (see Fig. 6-12). The area of the branch duct 
is 0.1 m
2
, and the area of the main duct is 0.2 m
2
 both upstram and downstream. What is the reduction in 
pressure between points u and d in the main duct? 
 
Solution: At 25 C, Table 6-2, ρ = 1.18425 kg/m3 
 Equation 6-17. 
 
( ) ddub2bd2ud2d AppcosAVAVAV −=βρ−ρ−ρ 
 β = 30o 
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 7 of 13 
 
sm/8
0.1
0.8
A
Q
V
b
b
b ===
 
 
sm/8
0.2
0.8-2.4
A
Q
V
u
u
u ===
 
 
s12 m/
0.2
2.4
A
Q
V
d
d
d ===
 
 
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )( )0.2pp30cos0.180.280.2121.18425 du222 −=−− 
 
 p
u
 - p
d
 = 62 Pa - - - - Ans. 
 
 
 
6-8. A two-branch duct system of circular duct is shown in Fig. 6-20. The fittings have the following equivalent 
length of straight duct: upstream to branch, 4 m; elbow, 2 m. There is negligible pressure loss in the straight-
through section of the branch. The designer selects 4 Pa/m as the pressure gradient in the 12- and 15-m 
straight sections. What diameter should be selected in the branch section to use the available pressure 
without dampering? 
 
 
Figure 6-20. Duct system in Prob. 6-8. 
 
 
 
 
 
 
 
Solution: 
 Available pressure drop 
 = ∆p = (12 m + 15 m)(4 Pa/m) = 108 Pa 
 Pressure gradient on 5 m and 6 m section. 
 
m7524
Pa108
L
p
+++
=
∆
 
 
Pa/m6
L
p
=
∆
 
 Figure 6-2, 6 Pa/m, 1.0 m
3
/s 
 
 D = 0.31 m - - - - Ans. 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 8 of 13 
 
6-9. A duct-system consists of a fan and a 25-m length of circular duct that delivers 0.8 m3/s of air. The installed 
cost is estimated to be $115 per square meter of sheet metal, the power cost is 6 cents per kilowatthour, the 
fan efficiency is 55 percent, and the motor efficiency 85 percent. There are 10,000 h of operation during the 
amortization period. Assume f = 0.02. What is the optimum diameter of the duct? 
 
Solution: Eq. 6-26. 
 
6
1
1
3
3
opt
C
HQ5C
D 





=
 
 Q = 0.8 m3/s 
 L = 25 m 
 H = 10,000 hrs 
 
 Eq. 6-20. 
 Initial Cost = (thickness)(piD)(L)(density of metal)(Installed cost / kg) 
 Initial Cost = (piD)(L)(Installed cost / m
2
) 
 Initial Cost = C
1
DL 
 
 C
1
 =(pi)(Installed cost / m
2
) 
 C
1
 = (pi)(115) = 361.3 
 
 Eq. 6-22. 
 Operating Cost = C
2
H∆pQ 
 C
2
 = [($0.06 / kwhr)(1 kw/1000 W)] / [(0.55)(0.85)] 
 C
2
 = 1.283422 x 10
-4 
 
 Eq. 6-23 
 
( )242
2
16
D
Q
D
L
fp
pi
ρ
=∆
 
 Eq. 6-24 
 
5
3
3
D
Q
LHCCostOperating =
 
 Substituting Eq. 6-23 to Eq. 6-22. 
 
( ) Q
16
D
Q
D
L
HfCCostOperating
242
2
2
pi
ρ
=
 
 
( ) 5
3
22
2
D
Q
LH
16
fC
CostOperating










pi
ρ
=
 
 
( )22 23
16
fC
C
pi
ρ
=
 
 Assume f = 0.02, ρ = 1.2041 kg/m3 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 9 of 13 
 
( )( )( )( )22
-4
3
16
1.20410.02101.283422
C
pi
×
=
 
 C
3
 = 8.122739x 10
-6 
 
 
6
1
1
3
3
opt
C
HQ5C
D 





=
 
 
( )( )( ) 6136-
opt
361.3
0.810000108.1227395
D 




 ×
=
 
 D
opt
 = 0.289 m - - - - Ans. 
 
6-10. Measurements made on a newly installed air-handling system were: 20 r/s fan speed, 2.4 m
3
/s airflow rate, 
340 Pa fan discharge pressure, and 1.8 kw supplied to the motor. These measurements were made with an 
air temperature of 20 C, and the system is eventually to operate with air at a temperature of 40 C. If the fan 
speed remains at 20 r/s, what will be the operating values of (a) airflow be the operating values of (a) airflow 
rate, (b) static pressure, and (c) power? 
 
Solution: At 20 C, 
 ω
1
 = 20 r/s 
 Q
1
 = 2.4 m
3
/s 
 SP
1
 = 340 Pa 
 P
1
 = 1.8 kw 
 ρ
1
 = 1.2041 kg/m
3
 
 
 At 40 C 
 ω
2
= 20 r/s 
 ρ
2
= 1.1272 kg/m
3
 
 
(a) Since ω is constant also Q is constant, 
 Q
2
 = 2.4 m
3
/s 
 
(b) Law 2, Q = constant 
 SP ~ ρ 
 
1
1
2
2 SPSP 





ρ
ρ
=
 
 
( )Pa 340
1.2041
1.1272
SP2 





=
 
 SP
2
 = 318 Pa - - - Ans. 
 
(c) Law 2, Q = constant 
 P ~ ρ 
 
1
1
2
2 PP 





ρ
ρ
=
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 10 of 13 
 
( )kw1.8
1.2041
1.1272
P2 





=
 
 P
2
 = 1.685 kw - - - Ans. 
 
6-11. A fan-duct system is designed so that when the air temperature is 20 C, the mass flow rate is 5.2 kg/s when 
the fan speed is 18 r/s and the fan motor requires 4.1 kw. A new set of requirement is imposed on the 
system. The operating air temperature is changed to 50 C, and the fan speed is increased so that the same 
mass flow of air prevails. What are the revised fan speed and power requirement? 
 
Solution: 
 At 20 C, Table 6-2 
 ρ
1
 = 1.2041 kg/m
3
 
 m
1
 = 5.2 kg/s 
 ω
1
 = 18 r/s 
 P
1
 = 4.1 kw 
 
 At 50 C, Table 6-2 
 ρ
2
 = 1.0924 kg/s 
 m2 = 5.2 kg/s 
 
 
/sm4.3186
1.2041
5.2m
Q 3
1
1
1 ==ρ
=
 
 
/sm4.7602
1.0924
5.2m
Q 3
2
2
2 ==ρ
=
 
 Revised fan speed, Equation 6-29 
 
 Q α ω or ω α 1/ρ 
 2
1
12 ρ
ρ
ω=ω
 
 
( ) ( )( )0924.1
2041.118=ω2
 
 ω
2
 = 19.84 r/s - - - Ans. 
 
 Revised power requirement, Equation 6-31. 
 
( )
2
PQV
SPQP
2
+=
 
 Equation 6-30 
 2
V
P
2ρ
α
 
 Then 
 2
QV
P
2ρ
α
 
 
2QαP
 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 11 of 13 
 
2
P
ρ
α
1
 
 
( )
2
2
2
2
11
2
1.0924
1.2041
4.1
P
P 





=
ρ
ρ
=
 
 P
2
 = 4.98 kw - - - Ans. 
 
6-12. An airflow rate of 0.05 m
3
/s issues from a circular opening in a wall. The centerline velocity of the jet is to be 
reduced to 0.75 m/s at a point 3 m from the wall. What should be the outlet velocity u
o
 of this jet? 
 
Solution: Equation 6-32. 
 
( )[ ]222
oo
x
r57.5x
A7.41u
u
+
=
1
 
 u = 0.75 m/s 
 x = 3 m 
 r = 0 
 x
A7.41u
u
oo
=
 
 o
o
u
Q
A =
 
 Q = 0.05 m
3
/s 
 
 x
Qu7.41
u
o
=
 
 
( )
3
0.05u7.41
0.75u
o
==
 
 u
o
 = 1.84 m/s - - - Ans. 
 
 
6-13. Section 6-19 points out that jets entrains air as they move away from their inlet into the room. The 
entrainment ratio is defined as the ration of the air in motion at a given distance x from the inlet to the airflow 
rate at the inlet Q
x
/Q
o
. Use the expression for the velocity in a circular jet, Eq. (6-32), multiplied by the area of 
an annular ring 2pirdr and integrate r from 0 to h to find the expression for Q
x
/Q
o
. 
 
Solution: Equarion 6-32. 
 
( )[ ]222
oo
x
r57.5x
A7.41u
u
+
=
1
 
 
( )∫∞ pi= 0x rdr2uQ 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 12 of 13 
 
( )[ ]∫
∞
+
pi=
0 2
2
2
oo
x
x
r57.51x
rdrA7.41u
2Q
 
 
oo
oo
oo
A
Q
A
Au
Au ==
 
 
( )[ ]∫
∞
+








pi=
0 2
2
2
oo
x
x
r57.51
rdr
Ax
7.41
2
Q
Q
 
 
( )[ ] rdr
x
r57.51
Ax
7.41
2pi
Q
Q 2
0
2
2
oo
x
−
∞
∫ +







=
 
 
 
 Let: 
 
( )22 xr57.51s += 
 
( )
rdr
x
57.52
ds
2
=
 
 Then: 
 
( ) ( )[ ] ( ) rdrx57.52xr57.51Ax57.52 7.41x2piQQ 220 22o
2
o
x 





+








=
−
∞
∫
 
 
∞
−












+
−
=
0
1
2
2
oo
x
x
r
57.51
A
0.405x
Q
Q
 
 
1)(0
A
0.405x
Q
Q
oo
x
−
−
=
 
 
Ans.- - - - 
A
0.405x
Q
Q
oo
x
=
 
 
6-14. From the equation for velocities in a plane jet, determine the total included angle between the planes where 
the velocities are one-half the centerline velocities at that x position. 
 
Solution: Equarion 6-33. 
 












−=
x
y
7.67tanh1
x
b2.40u
u 2o
 
 Centerline Velocity 
 y = 0 
 x
b2.40u
u oc =
 
 At 1/2 of centerline velocity at x position. 
CHAPTER 6 - FAN AND DUCT SYSTEMS 
Page 13 of 13 
 












−=
x
y
7.67tanh1
x
b2.40u
u 2oc2
1
 
 or 
 






−=
x
y
7.67tanh1
2
1 2
 
 
0.114912
x
y
=
 
 
 Total included angle , θ 
 
( )0.1149122Arctan
x
y
2Arctan =





=θ
 
 θ = 13.11o - - - Ans. 
 
 
- 0 0 0 - 
 
 
 
CHAPTER 7 - PIPING SYSTEMS 
Page 1 of 6 
7-1. A convector whose performance characteristics are shown in Fig. 7-4 is supplied with a flow rate of 0.04 kg/s 
of water at 90 C. The length of the convector is 4 m, and the room-air temperature is 18 C. What is the rate 
of heat transfer from the convector to the room air? 
 
Solution: 
 See Fig. 7-4 
 m = 0.04 kg/s 
 t
1
 = 90 C 
 L = 4 m 
 t
i
 = 18 C 
 
 p
12
mc
q
tt −=
 
 c
p
 = 4.19 kJ/kg.K 
 ( ) ( )( )41900.04
q
90t 2 −=
 
 
q0059666.090t 2 −= 
 Mean Water Temp. 
 
( )2121m ttt += 
 
( )0.0059666q-9090t
2
1
m += 
 
0.0029833q90tm −= 
 Equation for Fig. 7-4. 
 
W/m56016t
L
q
m −=
 
 
( )( )56016t4q m −= 
 
224064tq m −= 
 Substituting: 
 t
m
 = 90 - 0.0029833 (64t
m
 - 2240) 
 t
m
 = 81.182 C 
 
 q = 64t
m
 - 2,240 Watts 
 q = 64(81.182) - 2,240 Watts 
 q = 2,956 Watts 
 q = 2.956 kW ---- Ans. 
 
7-2. Compute the pressure drop in pascals per meter length when a flowrate of 8 L/s of 60 C water flows through 
a Schedule 40 steel pipe of nominal diameter 75 mm (a) using Eq. (7-1) and (b) using Figs. 7-6 and 7-7. 
 
Solution: 
 (a) Eq. 7-1. 
 
ρ=∆
2
V
D
L
fp
2
 
 
From Table 7-3 at 60 C. 
 ρ = 983.19 kg/m3 
 µ = 0.476 mPa.s = 0.000476 Pa.s 
 
 75-mm Schedule 40 Steel Pipe, Table 7-1, ID = 77.92 mm 
CHAPTER 7 - PIPING SYSTEMS 
Page 2 of 6 
 ( )
m/s1.678
4m0.07792
/sm0.008
V
2
3
=
pi
=
 
 Table 6-1, ε = 0.000046 commercial steel. 
 
0.00059
0.07792
0.000046
D
==
ε
 
 
( )( )( )
0.000476
983.190.077921.678DV
Re =
µ
pi
=
 
 Re = 270,067 
 
 From the Moody Chart, Fig. 6-1. 
 Re = 270,067, 
 
0.00059
D
=
ε
 
 f = 0.019 
 
ρ=∆
2
V
D
L
fp
2
 
 
( ) ( ) ( )983.19
2
1.678
0.07792
1
0.019
L
p
2






=
∆
 
 ∆∆∆∆p/L = 338 Pa/m ---- Ans.7-3. In the piping system shown schematically in Fig. 7-14 the common pipe has a nominal 75 mm diameter, the 
lower branch 35 mm, and the upper branch 50 mm. The pressure of water at the entrance is 50 kPa above 
atmospheric pressure, and both branches discharged to atmospheric pressure. The water temperature is 20 
C. What is the water flow rate in liters per second in each branch? 
 
 
 
 
Solution: 
 ∆p = 50 kPa - 0 - 50 kPa = 5000 Pa 
 Use Fig. 7-6, water temperature of 20 C 
 
 Table 7-4. 
 For 75-mm pipe 
 Elbow = 4 x 3 m = 12 m 
 Straight Pipe = 8 m + 4 m + 5 m + 7 m + 15 m = 39 m 
 L
1
 = 12 m + 39 m = 51 m 
 
CHAPTER 7 - PIPING SYSTEMS 
Page 3 of 6 
 For 50-mm pipe 
 Straight Branch = 0.9 m 
 Straight Pipe = 30 m 
 L
2
 = 0.9 m + 30 m = 30.9 m 
 
 For 35-mm pipe 
 Side Branch = 4.6 m 
 Straight Pipe = 6 m + 18 m = 24 m 
 Elbow = 1 x 1.2 m = 1.2 m 
 L
3
 = 4.6 m + 24 m + 1.2 m = 29.8 m 
 
 Q
1
 = Q
2
 + Q
3
 
 
 
pL
L
p
L
L
p
2
2
2
1
1
1 ∆=




 ∆
+




 ∆
 
 
pL
L
p
L
L
p
3
3
3
1
1
1 ∆=





 ∆
+




 ∆
 
 
32
2
2 L
L
p
L
L
p







 ∆
=




 ∆
3
3
 
 
( ) ( )29.8
L
p
30.9
L
p
2
2







 ∆
=




 ∆
3
3
 
 





 ∆
=






 ∆
2
2
L
p
1.036913
L
p
3
3
 
 Assume f = 0.02 
 r = 998.21 kg/m3. 
 For 75-mm pipe, ID = 77.92 mm = 0.07792 m 
 For 50-mm pipe, ID = 52.51 mm = 0.05251 m 
 For 35-mm pipe, ID = 35.04 mm = 0.03504 m 
 
 
ρ





=
∆
2
V
D
1
f
L
p 2
1 
 
VD
4
1
Q 2pi=
 
 
2D
4Q
V
pi
=
 
 
ρ








pi






=
∆
4
1
2
2
1
1
1
D
8Q
D
1
f
L
p
 
 
ρ








pi
=
∆
5
1
2
2
1
1
1
D
8Q
f
L
p
 
 
( ) ( )
2
152
2
1
1
1 5,633,748Q
0.07792
8Q
L
p
=








pi
=
∆
21.99802.0
 
 
( ) ( )
2
252
2
2
2
2 Q
0.05251
8Q
L
p
176,535,4021.99802.0 =








pi
=
∆
 
CHAPTER 7 - PIPING SYSTEMS 
Page 4 of 6 
 
( ) ( )
2
352
2
2
3
3 Q
0.03504
8Q
L
p 668,352,30621.99802.0 =








pi
=
∆
 
 
 
 (1) 
 
 
( ) ( ) 000,50=




 ∆
+




 ∆
30.9
L
p
51
L
p
2
2
1
1
 
 
( )( ) ( )( ) 50,00030.9Q40,535,176515,633,748Q 2221 =+ 
 
( )( ) ( )( ) 50,00030.9Q40,535,176515,633,748Q 2221 =+ 
 287,321,148Q
1
2
 + 1,253,093,138Q
2
2
 = 50,000 
 
 
 (2) 
 
 





 ∆
=






 ∆
2
2
L
p
1.036913
L
p
3
3
 
 
( )2223 Q40,535,1761.0369138Q306,352,66 = 
 Q
2
 = 2.7Q
3 
 
 Q
1
 = Q
2
 + Q
3
 
 Q
1
 = 2.7Q
3
 + Q
3
 
 Q
1
 = 3.7Q
3
 
 
 
 Then. 
 287,321,148Q
1
2
 + 1,253,093,138Q
2
2
 = 50,000 
 287,321,148(3.7Q
3
)
2
 + 1,253,093,138(2.7Q
3
)
2
 = 50,000 
 Q
3
 = 0.00196 m
3
/s 
 Q
3
 = 1.96 L/s - - - Ans. 
 Q
2
 = 2.7Q
3
 
 Q
2
 = 5.29 L/s - - - - Ans. 
 Q
1
 = 3.8Q
3
 
 Q
1
 = 7.25 L/s - - - - Ans. 
 
7-4. A centrifugal pump with the characteristics shown in Fig. 7-9 serves a piping network and delivers 10 L/s. An 
identical pump is placed in parallel with the original one to increase the flow rate. What is (a) the new flow 
rate in liters per second and (b) the total power required by the two pumps? 
 
Solution: 
 Use Fig. 7-9. 
 At 10 L/s. Pressure Rise, ∆p = 130 kPa 
 Efficiency = η = 62 % 
 P = (0.10)(130) / (0.62) = 2.097 kw 
 
 For the pipe network. 
 ∆p α Q2 
CHAPTER 7 - PIPING SYSTEMS 
Page 5 of 6 
 Q
1
 = 10 L/s 
 ∆p
1
 = 130 kPa 
 
 
 Use trial and error to find Q
2
 and ∆p
2
 that will lie along the pump curve in Fig. 7-9. 
 
 Trial 1, Q
2
 = 15 L/s 
 
 
2
1
2
12
Q
Q
pp 





∆=∆
 
 
( ) kPa292.5
10
15
130p
2
2 =





=∆
 
 Each pump = Q = 7.5 L/s 
 
 From Fig. 7-9. ∆p = 210 kPa < 292.5 kPa 
 Trial 2, Q
2
 = 13.3 L/s 
 
 
2
1
2
12
Q
Q
pp 





∆=∆
 
 
( ) kPa292.5
10
13.3
130p
2
2 =





=∆
 
 Each pump = Q = 6.65 L/s 
 
 From Fig. 7-9. ∆p = 230 kPa ~ 230 kPa 
 Efficiency = 80 % 
 Then Q
2
 = 13.3 L/s total - - - Ans. 
 Power = 2(0.00665)(230) / 0.80 
 Power = 3.82 kW - - - - Ans. 
 
7-5. An expansion tank is to be sized so that the change in air volume between the cold-water conditions (25 C) 
and the operating water temperature (85 C) is to be one fourth the tank volume. If pi = 101 kPa abs and p
c
 = 
180 kPa abs,., what will ph be? 
 
Solution: 
 Eq. 7-7. 
 
CB
t
h
i
b
i VV
V
p
p
p
p
1
−
=
−
 
 V
B
 - V
C
 = 0.25V
t
 
 
t
t
h
i
b
i 0.25V
V
p
p
p
p
1
=
−
 
 
0.25
p
p
p
p
h
i
c
i
=−
 
 
0.25
p
101
180
101
h
=−
 
 p
h
 = 325 kPa abs. - - - Ans. 
CHAPTER 7 - PIPING SYSTEMS 
Page 6 of 6 
 
- 0 0 0 - 
 
 
 
 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 1 of 6 
8-1. A cooling and dehumidifying coil is supplied with 2.4 m
3
/s of air at 29 C dry-bulb and 24 C wet-bulb 
temperatures, and its cooling capacity is 52 kW. The face velocity is 2.5 m/s. and the coil is of the direct-
expansion type provided with refrigerant evaporating at 7 C. The coil has an air-side heat-transfer area of 15 
m
2
 per square meter of face area per row of tubes. The ratio of the air-side to refrigerant-side area is 14. The 
values of h
r
 and h
c
 are 2050 and 65 W/m
2
.K, respectively. Calculate (a) the face area, (b) the enthalpy of 
outlet air, (c) the wetted-surface temperatures at the air inlet, air outlet, and at the point where the enthalpy 
of air is midway between its entering and leaving conditions, (d) the total surface area, (e) the number of 
rows of tubes, and (f) the outlet dry-bulb temperature of the air. 
 
Solution: 
 At 29 C dry-bulb and 24 C wet-bulb 
 h
a,1
=72.5 kJ/kg 
 g
a,1
 = 0.88 m
3
/kg 
 
 (a) Face Area = (2.4 m
3
/s) / (2.5 m/s) 
 Face Area = 0.96 m
2
 
 
 (b) Enthalpy of outlet air, h
a,2
 
 m = (2.4 m
3
/s) / (0.88 m
3
/kg) = 2.7273 kg/s 
 m
q
hh ta,1a,2 −=
 
 
kg/s2.7273
kW 52
kJ/kg72.5ha,2 −=
 
 h
a,2
 = 53.4 kJ/kg 
 
 (c) Wetted Surface Temperature 
 Eq. 8-1. 
 
( )ia
pm
c hh
c
dAh
dq −=
 
 Eq. 8-2. 
 
( )riir ttdAhdq −= 
 Eq. 8-3. 
 irpm
c
ia
ri
Ahc
Ah
hh
tt
R =
−
−
=
 
 t
r
 = 7 C 
 A/A
i
 = 14 
 h
r
 = 2050 W/m
2
.K 
 h
c
 = 65 W/m
2
.K 
 c
pm
 = 1.02 kJ/kg.K 
 
( )( )
( )( ) 0.435220501.02
1465
hh
tt
R
ia
ri
==
−
−
=
 
 h
a
 and h
i
 in kJ/kg 
 
 Eq. 8-4. 
 h
i
 = 9.3625+1.7861t
i
+0.01135t
i
2
+0.00098855t
i
3
 
 Eq. 8-5. 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 2 of 6 
 
0t0.000988550.01135t1.7861t9.3625h
R
t
R
t 3
i
2
iia
ri
=++++−−
 
 
At the air inlet, h
a,1
 = 72.5 kJ/kg 
 
0t0.000988550.01135t1.7861t9.362572.50.4352
7
04352
t 3
i
2
ii
i
=++++−−
 
 
By trial and error: t
i
 = 17.31 C and enthalpy h
i
 = 48.8 kJ/kg at air inlet. 
 
At the air outlet, h
a,3
 = 53.4 kJ/kg 
 
0t0.000988550.01135t1.7861t9.3625
0.4352
7
04352
t 3
i
2
ii
i
=++++−− 4.53
 
 
By trial and error: t
i
 = 13.6 C and enthalpy h
i
 = 38.23 kJ/kg at air outlet. 
 
At the midway enthalpy, h
a,2
 =(1/2)(72.5 kJ/kg + 53.4 kJ/kg) = 62.95 kJ/kg 
 
0t0.000988550.01135t1.7861t9.3625
0.4352
7
04352
t 3
i
2
ii
i
=++++−− 95.62
 
 
By trial and error: t
i
 = 15.5 C and enthalpy h
i
 = 43.46 kJ/kg at midway enthalpy. 
 
Answer - - - 17.31 C, 15.5 C, and 13.6 C. 
 
(d) Total surface area. 
 
 Between 1 and 2. 
 
( ) ( )differenceenthalpymean
c
Ah
hhmq
pm
21c
2121 −=−=
−
−
 
 c
pm
 = 1020 J/kg.K 
 
( )( ) ( ) 










 +





 +
=−
−
2
43.4648.8
-
2
62.9572.5
1020
A65
62.9572.52.7273 21
 
 A
1-2
 = 18.93 m
2
 
 
 Between 2 and 3. 
 
( ) ( )differenceenthalpymean
c
Ah
hhmq
pm
2c
322 −=−=
−
−
3
3
 
 c
pm
 = 1020 J/kg.K 
 
( )( ) ( ) 










 +





 +
=−
−
2
38.2343.46
-
2
62.95
1020
A65
62.952.7273 2
4.534.53 3
 
 A
2-3
 = 23.59 m
2
 
 
 Surface Area of Coil = 18.93 m
2
 + 23.59 m
2
 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 3 of 6 
 Surface Area of Coil = 42.52 m
2
 - - - Ans. 
 
 (e) The number of rows of tubes. 
 
 No. of rows = (42.52 m
2
)/[(15 m
2
/m
2
)(0.96 m
2
)] 
 No. of rows = 3 rows - - - Ans. 
 
 (f) Outlet dry-bulb temperature. 
 
 Q
s
 = (2.7273 kg/s)(c
pm
)(t
i
 - t
2
) 
 c
pm
 = 1020 J/kg.K 
 
 







 +
−
+
=
−
2
tt
2
tt
hAQ
i,2i,121
c21s
 
 Between 1 and 2. 
 
 
 
( )( )( ) ( )( ) 




 +
−
+
=−
2
15.517.31
2
t29
6518.93t2910202.7273 22
 
 t
2
 = 23.75 C 
 
 Between 2 and 3. 
 
 
 
( )( )( ) ( )( ) 




 +
−
+
=−
2
13.615.5
2
t23.75
6523.59t23.7510202.7273 23
 
 t
3
 = 19.8 C - - - Ans. 
 
8-2. For the area A1-2 in Example 8-2 using the entering conditions of the air and the wetted-surface 
temperatures at points 1 and 2, (a) calculate the humidity ratio of the air at point 2 using Eq. (8-6), and (b) 
check the answer with the humidity ratio determined from the dry-bulb temperature and enthalpy at point 2 
calculated in Example 8-1. 
 
Solution: (a) See Example 8-2. 
 
 Entering Conditions at Point 1 
 h
a
 = 60.6 kJ/kg 
 t
r
 = 12.0 F 
 t
i
 = 16.28 F 
 h
i
 = 45.72 kJ/kg 
 
 Wetted-Surface at point 2 
 h
a
 = 48.66 kJ/kg 
 t
r
 = 9.5 F 
 t
i
 = 12.97 F 
 h
i
 = 36.59 kJ/kg 
 
 Eq. 8-6. 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 4 of 6 
 
( ) 





 +
−
+
=−
−
2
WW
2
WW
c
Ah
WWG
i,2i,121
pm
21c
21
 
 G = 2.5 kg/s 
 A
1-2
 = 41.1 m2 
 c
pm
 = 1.02 kJ/kg.K = 1020 W/kg.K 
 h
c
 = 55 W/m
2
.K 
 
 For W1, psychrometric chart 
 At 30 C dry-bulb and 21 C wet-bulb temperature. 
 W
1
 = 0.012 kg/kg 
 Table A-2. 
 t
i,1
 = 16.28 C, 
 W
i,1
 = 0.01163 kg/kg 
 
 t
i,2
 = 12.97 C 
 W
i,2
 = 0.00935 kg/kg 
 
 Solve for W
2
 by substituing to Eq. 8-6. 
 
( )( ) ( )( ) 




 +
−
+
=−
2
0.009350.01163
2
W0.012
1020
41.155
W0.0122.5 22
 
 W
2
 = 0.0111 kg/kg - - - Ans. 
 
 (b) Checking W
2
: 
 
 At point 2., h
a,2
 = 48.66 F, t
2
 = 20.56 C 
 
 From psychrometric chart, Figure 3-1 
 
 W
2
 = 0.011 kg/kg - - - Ans. 
 
8-3. A direct-expansion coil cools 0.53 kg/s of air from an entering condition of 32 C dry-bulb and 20 C wet-bulb 
temperature. The refrigerant temperature is 9 C, h
r
 = 2 kW/m
2
.K, h
c
 = 54 W/m
2
.K, and the ratio of air-side to 
refrigerant-side areas is 15. Calculate (a) the dry-bulb temperature of the air at which condensation begins 
and (b) the surface area in square meters of the portion of the coil that is dry. 
 
 
Solution: 
 
 m = 0.53 kg/s 
 At 32 C dry-bulb and 20 C wet-bulb temperatures 
 h
a,1
 = 57 kJ/kg 
 
 (a) Dew-point of entering air = t
i,2
 = 13.8 C 
 
 Equation 8-11. 
 
( ) ( )
A
dAAh
ttdAhtt irri,2ci,22 −=−
 
 Then: 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 5 of 6 
 
( ) ( )
A
Ah
tthtt irri,2ci,22 −=−
 
 h
c
 = 54 W/m2.K 
 h
r
 = 2000 W/m2.K 
 t
r
 = 9 C 
 A/A
i
 = 15 
 
( )( ) ( )( )
15
20009-13.8
5413.8t 2 =−
 
 t
2
 = 25.7 C - - - - Ans. 
 
 (b) 
 
 
( ) 





−
+
+
=−
− r
21
21
ric
21pm t
2
tt
A
hA
A
h
1
1
ttGc
 
 c
pm
 = 1020 J/kg.K 
 G = 0.53 kg/s 
 
( )( )( ) 





−
+
+
=−
−
9
2
25.735
A
2000
15
54
1
1
25.73210200.53 21
 
 A
1-2
 = 4.47 m2 - - - Ans. 
 
 
8-4. For a coil whose performance and conditions of entering air are shown in Table 8-1, when the face velocity is 
2 m/s and the refrigerant temperature is 4.4 C, calculate (a) the ratio of moisture removal to reduction in dry-
bulb temperature in the first two rows of tubes in the direction of air flow in the last two rows and (b) the 
average cooling capacity of the first two and the last two rows in kilowatts per square meter of face area. 
 
Solution: Use Table 8-1. 
 Face velocity = 2 m/s 
 Refrigerant Temperature = 4.4 C. 
 
 (a) First 2-rows: 
 At 30 C dry-bulb, 21.7 C wet-bulb temperature 
 h
1
 = 63 kJ/kg 
 W
1
 = 0.013 kg/kg 
 γ
1
 = 0.08735 m
3
/kg 
 Final DBT = 18.2 C 
 Final WBT = 17.1 C 
 h
2
 = 48.5 kJ/kg 
 W
2
 = 0.0119 kg/kg 
 
 Ratio for the first two rows = (W
1
 - W
2
) / (t
1
 - t
2
) 
 = (0.013 - 0.0119) / (30 - 18.2) 
 = 0.0000932 kg/kg.K - - - Ans. 
 
 For the last two rows. 
 Rows of tube = 4 in Table 8-1. 
 Final DBT = 14.3 C 
 Final WBT = 13.8 C 
CHAPTER 8 - COOLING AND DEHUMIDIFYING COILS 
 Page 6 of 6 
 h
3
 = 38.5 kJ.kg 
 W
3
 = 0.0095 kg/kg 
 Ratio for the last two rows = (W
2
 - W
3
) / (t
2
 - t
3
) 
 = (0.0119 - 0.0095) / (18.2 - 13.8) 
 = 0.00055 kg/kg.K - - - Ans. 
 
 (b) First two rows. 
 kW per sq m of face area 
 = [(2 m/s)/(0.8735 m
3
/kg)](h
1
 - h
2
) 
 = (2 / 0.8735)(63 - 48.5) 
 = 33.2 kw - - - Ans. 
 For the last two rows. 
 kW per sq m of face area 
 = [(2 m/s)/(0.8735 m
3
/kg)](h
2
 - h
3
) 
 = (2 / 0.8735)(48.5 - 38.5) 
 = 22.9 kw - - - Ans. 
 
8-5. An airflow rate of 0.4 kg/s enters a cooling and dehumidifying coil, which for purpose of analysis is divided 
into two equal areas, A
1-2
 and A
2-3
. The temperatures of the wetted coil surfaces are t
i,1
 = 12.8 C, t
i,2
 = 10.8 C 
, and t
i,3
 = 9.2 C. The enthalpy of entering air h
a,1
 = 81.0 and 
ha,2
 = 64.5 kJ/kg. Determine 
ha,3
. 
 
Soution: 
 G = 0.4 kg/s 
 Then equation: 
 















 +
−






 +
=















 +
−






 +
−−
2
hh
2
hh
q
2
hh
2
hh
q
i,3i,2a,3a,2
21
i,2i,1a,2a,1
21
 
 Eq. 8-4. 
 
3
i
2
iii t0.000988550.01135t1.786t9.3625h +++= 
 At t
i,1
 = 12.8 C 
 h
i,1
 = 36.16 kJ/kgAt t
i,2
 = 10.8 C 
 h
i,2
 = 31.22 kJ/kg 
 At t
i,3
 = 9.2 C 
 h
i,3
 = 27.52 kJ/kg 
 Then: 
 















 +
−






 +
=















 +
−






 +
2
hh
2
hh
h- h
2
hh
2
hh
h-h
i,3i,2a,3a,2
a,3a,2
i,2i,1a,2a,1
a,2a,1
 
 













 +
−






 +
=











 +
−




 +
2
27.5231.22
2
h64.5
h- 64.5
2
31.2236.16
2
64.581
64.5-81
a,3
a,3
 
 0.422427(0.5h
a,3
 + 2.88) = 64.5 - h
a,3
 
 1.211214h
a,3
 = 63.28341 
 h
a,3
 = 52.25 kJ/kg - - - Ans. 
 
- 0 0 0 - 
CHAPTER 9 - AIR-CONDITIONING CONTROLS 
 Page 1 of 4 
9-1. A space thermostat regulates the damper in the cool-air supply duct and thus provides a variable air flow 
rate. Specify whether the damper should be normally open or normally closed and whether the thermostat is 
direct- or reverse-acting. 
 
Answer: Use normally closed damper and reverse-acting thermostat since as the space temperature increases the 
volume rate of air will increase the pressure will reduce. 
 
9-2. On the outdoor-air control system of Example 9-4, add the necessary features to close the outdoor-air 
damper to the minimum position when the outdoor temperature rises above 24 C. 
 
Answer: Add a diverting relay. Pressure will divert to 68 kPa (20 %) minimum position when the outdoor temperature 
rises above 24 C. 
 
 
 
 
 
9-3. The temperature transmitter in an air-temperature controller has a range of 8 to 30 C through which range the 
pressure output change from 20 to 100 kPa. If the gain of the receiver-controller is set at 2 to 1 and the 
spring range of the cooling-water valve the controller regulates is 28 to 55 kPa, what is the throttling range of 
this control? 
 
Solution: 
 Output of temperature transmitter = (100 -20 kPa) / (30 - 8 C) 
 = 3.6364 kPa/K 
 
 Throttling Range = (55 kPa - 28 kPa) / [(2)(3.6364 kPa/K)] 
 Throttling Range = 3.7 K . . . Ans. 
 
9-4 The air supply for a laboratory (Fig. 9-29) consists of a preheat coil, humifidier, cooling coil, and heating coil. 
The space is to be maintained at 24 C, 50 percent relative humidity the year round, while the outdoor supply 
air may vary in relative humidity between 10 and 60 percent and the temperature from -10 to 35 C. The 
spring ranges available for the valves are 28 to 55 and 62 to 90 kPa. Draw the control diagram, adding any 
additional components needed, specify the action of the thermostat(s) and humidistat, the spring ranges of 
the valves, and whethet they are normally open or n ormally closed. 
 
 
 
Answer: 
 
Spring range = 28 to 55 kPa 
 = 62 to 90 kPa 
 
(a) Limitiations: 
 
CHAPTER 9 - AIR-CONDITIONING CONTROLS 
 Page 2 of 4 
 Use preheat coil when space temperature is less than 24 C. 
 Use humidifier when space relative humidity is less than 50 percent. 
 Use cooling coil when space temperature is greater than 24 C. 
 Use reheat coil when space relative humidity is greater than 5- percent. 
 
(b) Valves. 
 
 Preheat coil and reheat coil has normally open valves. 
 Humidifier has normally closed valves. 
 Cooling coils has normally closed valves. 
 
(c) Action 
 Action of thermostat is to control temperature by preheat coil and cooling coil. 
 Action of humidistat is to control temperature by humidifier and reheat coil. 
 
(d) Control diagram 
 
 
 
 
 
 
9-5. A face-and-bypass damper assembly at a cooling coil is sometimes used in humid climates to achieve 
greater dehumidification for a given amount of sensible cooling, instead of permitting all the air to pass over 
the cooling coil. Given the hardware in Fig. 9-30, arrange the control system to regulate the temperature at 
24 C and the relative humidity at 50 percent. If both the temperature and humidity cannot be maintained 
simultaneously, the temperature control should override the humidity control. The spring ranges available for 
the valve and damper are 28 to 55 and 48 to 76 kPa. Draw the control diagram and specify the action of the 
thermostat and humidistat, whether the valve is normally open or normally closed, and which damper is 
normally closed. 
CHAPTER 9 - AIR-CONDITIONING CONTROLS 
 Page 3 of 4 
 
 
Answer: 
 
 By-pass damper is normally open, 
 Face damper is normally open. 
 Chilled water valve is normally open. 
 Damper operator is normally open 
 
 Chilled water valve opens when space temperature is greater than 24 C. 
 Chilled water valve closes when space temperature is less than 24 C. 
 
 Damper operator is closing when the relative humidity is greater than 50 percent. 
 Damper operator is opening when the relative humidity is less than 50 percent. 
 Damper operator is opening the by-pass damper while closing the face damper from N.O. position. 
 
Diagrams.
 
 
 When temperatue control override the humidity control. 
 1. Chilled water valve closing when temperature is less than 24 C and opeing when temperature is greater 
than 24 C. 
 
CHAPTER 9 - AIR-CONDITIONING CONTROLS 
 Page 4 of 4 
 2. Damper is closing when the temperature is less than 24 C and opening when temperature is greater than 
24 C. 
 
 
 
9-6. Section 9-18 described the flow characteristics of a coil regulated by a valve with linear characteristics. The 
equation of the flow-steam position for another type of valve mentioned in Sec. 9-18, the equal-percentage 
valve, is 
 
 
1
100
strokestemofpercent
xwhereA
pC
Q x
v
−==
∆
 
 
 If such a valve with an A value of 20 and a Cv of 1.2 is applied to controlling the coil in Fig. 9-25 with Dcoil = 
2.5Q2 and the total pressure drop across the valve and coil of 80 kPa, what is the flowrate when the valve 
stem stroke is at the halfway position? (Compare with a linear-characteristic valve in Fig. 9-27.) 
 
Solution: 
 
22.5QkPa80p −=∆
 
 C
v
 = 1.2 
 A = 20 
 at percent of stem stroke = 50 %. 
 
0.51
100
50
x −=−=
 
 
x
v
A
pC
Q
=
∆
 
 
0.5-
2
20
2.5Q801.2
Q
=
− 
 
22.5Q800.26833Q −= 
 Q = 2.21 L/s - - - Ans. 
 
 Comparing to linear-characteristic valve in Fig. 9-27. 
 
pC
100
strokepercent
Q v ∆=
 
 
( ) 22.5Q801.2
100
50
Q −=
 
 Q = 3.893 L/s > 2.21 L/s 
 
- 0 0 0 - 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 1 of 10 
10-1. A Carnot refrigeration cycle absorbs heat at -12 C and rejects it at 40 C. 
(a) Calculate the coefficient of performance of this refrigeration cycle. 
(b) If the cycle is absorbing 15 kW at the -12 C temperature, how much power is required? 
(c) If a Carnot heat pump operates between the same temperatures as the above refrigeration cycle, 
what is the performance factor? 
(d) What is the rate of heat rejection at the 40 C temperature if the heat pump absorbs 15 kW at the -
12 C temperature? 
 
Solution: 
 
 (a) Coefficient of performance = T
1
 / (T
2
 - T
1
) 
 T
1
 = -12 C + 273 = 261 K 
 T
2
 = 40 C + 273 = 313 K 
 Coefficient of performance = 261 / (261 + 313) 
 Coefficient of performance = 5.02 - - - Ans. 
 (b) Coefficient of performance = useful refrigeration / net work 
 5.02 = 15 kw / net work 
 net work = 2.988 kW - - - Ans. 
 (c) Performance factor = coefficient of performance + 1 
 Performance factor = 6.01 - - - Ans. 
 (d) Performance factor=heat rejected from cycle/work required. 
 
 
15kwrejectedheat
rejectedheat
factorePerformanc
−
=
 
 
15kwrejectedheat
rejectedheat
6.02
−
=
 
 Heat rejected = 17.988 kw - - - Ans.10-2. If in a standard vapor-compression cycle using refrigerant 22 the evaporating temperature is -5 C and the 
condensing temperature is 30 C, sketch the cycle on pressure-enthalpy coordinates and calculate (a) the 
work of compression, (b) the refrigerating effect, and (c) the heat rejected in the condenser, all in kilojoules 
per kilograms , and (d) the coefficient of performance. 
 
Solution. 
 
 
 
At pont 1, Table A-6, -5 C, 
h
1
 = 403.496 kJ/kg 
s
1
 = 1.75928 kJ/kg.K 
 
At point 2, 30 C condensing temperature, constant entropy, Table A-7. 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 2 of 10 
h
2
 = 429.438 kJ/kg 
 
At point 3, Table A-6, 30 C 
h
3
 = 236.664 kJ/kg 
 
h
4
 = h
3
 = 236.664 kJ/kg 
 
(a) Work of compression = h
2
 - h
1
 
 = 429.438 - 403.496 
 = 25.942 kJ/kg - - - Ans. 
 
(b) Refrigerating effect = h
1
 - h
4
 
 = 403.496 - 236.664 
 = 166.832 kJ/kg - - - Ans. 
 
(c) Heat rejected = h
2
 - h
3
 
 = 429.438 - 236.664 
 = 192.774 kJ/kg - - - Ans. 
 
(d) Coefficient of performance 
 
 12
41
hh
hh
eperformancoftCoefficien
−
−
=
 
 
496.403438.429
664.236496.403
−
−
=eperformancoftCoefficien
 
 Coefficient of performance = 6.43 - - - Ans. 
 
10-3. A refrigeration system using refrigerant 22 is to have a refrigerating capacity of 80 kw. The cycle is a 
standard vapor-compression cycle in which the evaporating temperature is -8 C and the condensing 
temperature is 42 C. 
(a) Determine the volume flow of refrigerant measured in cubic meter per second at the inlet to the 
compressor. 
(b) Calculate the power required by the compressor. 
(c) At the entrance to the evaporator what is the fraction of vapor in the mixture expressed both on a 
mass basis and a volume basis? 
Solution: 
 
 
At 1, Table A-6, -8 C. 
 h
1
 = h
g1
 = 402.341 kJ/kg 
 h
f1
 = 190.718 kJ/kg 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 3 of 10 
 ν
g1
 = 61.0958 L/kg 
 ν
f1
 = 0.76253 L/kg 
 s
1
 = 1.76394 kJ/kg.K 
 
At 2, 42 C condensing temperature, constant entropy, Table A-7. 
 
 h
2
 = 438.790 kJ/kg 
 
At 3, Table A-6, 42 C 
 
 h
3
 = 252.352 kJ/kg 
 
h
4
 = h
3
 = 252.352 kJ/kg 
 
(a) Volume flow of refrigerant = wν
g 
 w(h
1
 - h
4
) = 80 kw 
 w (402.341 - 252.352) = 80 
 w = 0.5334 kg/s 
 
 Volume flow of refrigerant 
 = (0.5334 kg/s)(61.0958 L/kg) 
 = 32.59 L/s 
 = 0.03259 m
3
/s - - - Ans. 
 
(b) Power required by compressor 
 = w(h2 - h1) 
 = (0.5334)(438.790 - 402.341) 
 = 19.442 kw - - - Ans. 
 
(c) Let x
m
 = fraction of vapor by mass basis and 
 x
v
 = fraction of vapor by volume basis. 
 
 Mass Basis: 
 
190.718402.341
190.718252.352
hh
hh
x
f1g1
f14
m
−
−
=
−
−
=
 
 x
m
 = 0.292 - - - Ans. 
 
 Volume Basis: 
 Total volume = (1 - 0.292)(0.76253) + 0.292(61.0958) = 18.38 L/s 
 
( )
18.38
61.09580.292
x v =
 
 x
v
 = 0.971 - - - Ans. 
 
10-4. Compare the coefficient of performance of a refrigeration cycle which uses wet compression with that of one 
which uses dry compression. In both cases use ammonia as the refrigerant, a condensing temperature of 30 
C, and an evaporating temperature of -20 C; assume that the compressors are isentropic and that the liquid 
leaving the condenser is saturated. In the wet-compression cycle the refrigerant enters the compressor in 
such a condition that it is saturated vapor upon leaving the compressor. 
 
Solution: 
 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 4 of 10 
 For Dry Compression: 
 
 
 
At 1, -20 C, Table A-3. 
 h
1
 = h
g
 = 1437.23 kJ/kg 
 h
f
 = 108.599 kJ/kg 
 s
1
 = s
g
 = 5.9025 kJ/kg.K 
 s
f
 = 0.65436 kJ/kg.K 
 
At 2, 30 C Condensing Temperature, constant entropy, Fig. A-1. 
 h
2
 = 1704 kJ/kg 
 
At 3, 30 C, Table A-3. 
 h
3
 = 341.769 kJ/kg 
 s
3
 = 1.48762 kJ/kg.K 
 
At 4, s
4
 = s
3
, 
 
0.1588
0.654365.9025
0.654361.48762
ss
ss
x
fg
f4
=
−
−
=
−
−
=
 
 h
4
 = h
f
 + x(h
g
 - h
f
) 
 h
4
 = 108.599 + (0.1588)(1437.23 - 108.599) = 319.586 kJ/kg 
 
1437.231704
319.5861437.23
hh
hh
eperformancoftCoefficien
12
41
−
−
=
−
−
=
 
 Coefficient of performance = 4.19 
 
 For Wet Compression: 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 5 of 10 
 
 
At 2, 30 C condensing temperature, saturated, Table A-3. 
 h
2
 = 1486.14 kJ/kg 
 s
2
 = 5.2624 kJ/kg.K 
 
At 1, s
1
 = s
2
. 
 
0.878
0.654365.9025
0.654365.2624
ss
ss
x
fg
f1
=
−
−
=
−
−
=
 
 h
1
 = h
f
 + x (h
g
 - h
f
) 
 h
1
 = 108.599 + (0.878)(1437.23 - 108.599) 
 h
1
 = 1275.14 kJ/kg 
 
 h
3
 = 341.769 kJ/kg 
 h
4
 = 319.586 kJ/kg 
 
 
1275.14-1486.14
319.5861275.14
hh
hh
eperformancoftCoefficien
12
41 −
=
−
−
=
 
 
 Coefficient of performance = 4.53 
 
 Ans. 4.53 wet versus 4.19 dry. 
 
 
10-5. In the vapor-compression cycle a throttling device is used almost universally to reduce the pressure of the 
liquid refrigerant. 
(a) Determine the percent saving in net work of the cycle per kilograms of refrigerant if an expansion 
engine would be used to expand saturated liquid refrigerant 22 isentropically from 35 C to the 
evaporator temperature of 0 C. Assume that compression is isentropic from saturated vapor at 0 C 
to a condenser pressure corresponding yo 35 C. 
(b) Calculate the increase in refrigerating effect in kilojoules per kilograms resulting from use of 
expansion engine. 
 
Solution: 
 
 
 Vapor-Compression Cycle: 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 6 of 10 
 
 
 At 1, 0 C, Table A-6. 
 h
1
 = 405.361 kJ/kg 
 s
1
 = s
g1
 = 1.75279 kJ/kg.K 
 
 At 2, 35 C, constant entropy, Table A-7. 
 h
2
 = 430.504 kJ/kg 
 
 At 3, Table A-6 
 h
3
 = 243.114 kJ/kg 
 
 h
4
 = h
3
 = 243.114 kJ/kg 
 
 Net Work = h
2
 - h
1
 = 430.504 - 405.361 = 25.143 kJ/kg 
 Refrigerating Effect = h
1
 - h
4
 = 405.361 - 243.114 = 162.247 kJ/kg 
 
 
 For expansion engine: 
 
 
 
 At a, 0 C, Table A-6. 
 h
a
 = h
ga
 = 405.361 kJ.kg 
 h
fa
 = 200 kJ/kg 
 s
a
 = s
ga
 = 1.75279 kJ/kg.K 
 s
fa
 = 1.00000 kJ/kg.k 
 
 At b, constant entropy, Table A-2 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 7 of 10 
 h
b
 = 430.504 kJ/kg 
 
 At c, Table A-6. 
 h
c
 = 243.114 kJ/kg 
 s
c
 = 1.14594 kJ/kg 
 
 At d, constant entropy. 
 
0.193866
1.000001.75279
1.000001.14594
ss
ss
x
faga
fad
=
−
−
=
−
−
=
 
 h
d
 = h
fa
 + x(h
ga
 - h
fa
) 
 h
d
 = 200 + (0.193866)(405.361 - 200) 
 h
d
 = 239.813 kJ/kg 
 
 Net Work = (h
b
 - h
a
) - (h
c
 - h
d
) 
 Net Work = (430.5 - 405.361) - (243.114 - 239.813) 
 Net Work = 21.838 kJ/kg 
 Refrigerating Effect = h
a
 - h
d
 = 405.361 - 239.813 = 165.548 kJ/kg 
 
 (a) Percent Saving 
 
( )100%
25.143
21.83825.143 −
=
 
 = 13.1 % - - - Ans. 
 (b) Increase in refrigerating effect. 
 = 165.548 kJ/kg - 162.247 kJ/kg 
 = 3.301 kJ/kg - - - Ans. 
 
10-6. Since a refrigeration system operates more efficiently when the condensing temperature is low, evaluate the 
possibility of cooling the condenser cooling water of the refrigeration system in question with another 
refrigeration system. Will the compressor performance of the two systems be better, the same, or worse than 
one individual system? Explain why. 
 
Solution:CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 8 of 10 
 
 
 Coefficient of performance of two system: 
 
)h(hw)h(hw
)h(hw)h(hw
COP
aba121
daa411
c
−+−
−+−
=
 
 Coefficient of performance of each system 
 
)h(hw
)h(hw
COP
121
411
1
−
−
=
 
 
)h(hw
)h(hw
COP
aba
daa
2
−
−
=
 
 Substituting: 
 
 2
daa
1
411
daa411
c
COP
)h(hw
COP
)h(hw
)h(hw)h(hw
COP
−
+
−
−+−
=
 
 
 if COP
1
 = COP
2
 then: 
 
 COP
c
 = COP
1
 = COP
2
 
 
 Therefore it is the same COP as for individual system having equal COP and in between if COP is not 
the same..Ans. 
 
 
10-7. A refrigerant 22 vapor compression system includes a liquid-to-suction heat exchanger in the system. The 
heat exchanger warms saturated vapor coming from the evaporator from -10 to 5 C with liquid which comes 
from the condenser at 30 C. The compressions are isentropic in both cases listed below. 
(a) Calculate the coefficient of performance of the system without the heat exchanger but with the 
condensing temperature at 30 C and an evaporating temperature of -10 C. 
(b) Calculate the coefficient of performance of the system with the heat exchanger? 
(c) If the compressor is capable of pumping 12.0 L/s measured at the compressor suction, what is the 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 9 of 10 
refrigeration capacity of the system without the heat exchanger? 
(d) With the same compressor capacity as in (c), what is the refrigerating capacity of the system with 
the heat exchanger? 
 
Solution: 
 
 
 (a) Without heat exchanger 
 At 1,6, -10 C, Table A-6. 
 h
1
 = 401.555 kJ/kg 
 s
1
 = 1.76713 kJ/kg.K 
 At 2, 30 C, constant entropy, Table A-7 
 h
2
 = 431.787 kJ/kg 
 At 3,4 , 30 C, Table A-6. 
 h
3
 = 236.664 kJ/kg 
 At 5, h
5
 = h
3
 = 236.664 kJ/kg 
 
 
401.555431.787
236.664401.555
hh
hh
eperformancoftcoefficien
12
51
−
−
=
−
−
=
 
 coefficient of performance = 5.46 . . . Ans. 
 
 (b) With heat exchanger 
 At 6, -10 C , Table A-6 
 h
6
 = 401.555 kJ/kg 
 At 1, -10 C evaporator temperature, 5 C, Table A-7 
 h
1
 = 411.845 kJ/kg 
 At 2, 30 C, constant entropy, Table A-7 
 h
2
 = 444.407 kJ/kg 
 At 3, 30 C, table A-6 
 h
3
 = 236.664 kJ/kg. 
 
 Since no mention of subcooling. 
 h
5
 = h
4
 = h
3
 = 236.664 kJ/kg 
 
 
411.845444.407
236.664411.845
hh
hh
eperformancoftcoefficien
12
51
−
−
=
−
−
=
 
 coefficient of performance = 5.38 . . . Ans. 
 
CHAPTER 10 - THE VAPOR-COMPRESSION CYCLE 
 Page 10 of 10 
 (c) Refrigerating capacity without heat exchanger 
 At 1, ν = 65.3399 L/kg 
 
 Refrigerating Capacity 
 
( )51 hh
L/kg65.3399
L/s12.0
−





=
 
 
( )664.236−





= 401.555
L/kg65.3399
L/s12.0
 
 = 30.3 kW - - - - Ans. 
 
 
 (d) Refrigerating capacity with heat exchanger 
 At 1, ν = 70.2751 L/kg 
 
 Refrigerating Capacity 
 
( )51 hh
L/kg70.2751
L/s12.0
−





=
 
 
( )664.236−





= 411.845
L/kg70.2751
L/s12.0
 
 = 29.9 kW - - - - Ans. 
 
- 0 0 0 - 
CHAPTER 11 - COMPRESSORS 
Page 1 of 6 
11-1. An ammonia compressor has a 5 percent clearance volume and a displacement rate of 80 L/s and pumps 
against a condensing temperature of 40 C. For the two different evaporating temperatures of -10 and 10 C, 
compute the refrigerant flow rate assuming that the clearance volumetric efficiency applies. 
 
Solution: 
 Equation 11-7. 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
 (a) At -10 C, Table A-3. 
 s
1
 = 5.7550 kJ/kg 
 ν
suc
 = 417.477 L/kg 
 
 At 40 C, constant entropy, Fig. A-1 
 ν
dis
 = 112.5 L/kg 
 
 m = 5 % 
 
 Equation 11-4 and Equation 11-5. 
 
 








−
ν
ν
−= 1m100η
dis
suc
vc
 
 
 
86.4451
112.5
417.477
5100ηvc =





−−=
 
 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
417.477
100
86.445
L/s80w ×=
 
 w = 0.166 kg/s at -10 C - - - Ans. 
 (b) At 10 C, Table A-3 
 s1 = 5.4924 kJ/kg.K 
 ν
suc
 = 205.22 L/kg 
 
 At 40 C, constant entropy, Fig. A-1 
 ν
dis
 = 95 L/kg 
 
 m = 5 % 
 
 Equation 11-4 and Equation 11-5. 
 
 








−
ν
ν
−= 1m100η
dis
suc
vc
 
 
 
94.1991
95
205.22
5100ηvc =





−−=
 
 
CHAPTER 11 - COMPRESSORS 
Page 2 of 6 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
205.22
100
94.199
L/s80w ×=
 
 w = 0.367 kg/s at 10 C - - - Ans. 
 
11-2. A refrigerant 22 compressor with a displacement rate of 60 L/s operates in a refrigeration system that 
maintains a constant condensing temperature of 30 C. Compute and plot the power requirement of this 
compressor at evaporating temperatures of -20, -10, 0, 10 and 20 C. Use the actual volumetric efficiencies 
from Fig. 11-12 and the following isentropic works of compression for the five evaporating temperatures, 
respectively, 39.9, 30.2, 21.5, 13.7, and 6.5 kJ/kg. 
 
 
Solution: 
 (a) At -20 C evaporating temperature, Table A-6. 
 ν
suc
 = 92.8432 L/kg 
 p
suc
 = 244.83 kPa 
 Table A-7, 30 C 
 p
dis
 = 1191.9 kPa 
 
 Ratio = p
dis
 / p
suc
 = 1191.9 kPa / 244.82 kPa = 4.87 
 Figure 11-12 
 η
va
 = volumetric efficiency = 67.5 % 
 suc
va
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
92.8432
100
67.5
L/s60w ×=
 
 w = 0.4362 kg/s at -20 C 
 
 P = w∆h
i
 
 ∆h
i
 = 39.9 kJ/kg 
 P = (0.4362)(39.9) 
 P = 17.4 kw at -20 C 
 
 (b) At -10 C evaporating temperature, Table A-6. 
 ν
suc
 = 65.3399 L/kg 
 p
suc
 = 354.3 kPa 
 Table A-7, 30 C 
 p
dis
 = 1191.9 kPa 
 
 Ratio = p
dis
 / p
suc
 = 1191.9 kPa / 354.30 kPa = 3.364 
 Figure 11-12 
 η
va
 = volumetric efficiency = 77.5 % 
 suc
va
100ratentdisplacemew
ν
η
×=
 
CHAPTER 11 - COMPRESSORS 
Page 3 of 6 
 
( ) ( )
65.3399
100
77.5
L/s60w ×=
 
 w = 0.7117 kg/s at -10 C 
 
 P = w∆h
i
 
 ∆h
i
 =30.2 kJ/kg 
 P = (0.7117)(30.2) 
 P = 21.5 kw at -10 C 
 
 (c) At 0 C evaporating temperature, Table A-6. 
 ν
suc
 = 47.1354 L/kg 
 p
suc
 = 497.59 kPa 
 Table A-7, 30 C 
 p
dis
 = 1191.9 kPa 
 
 Ratio = p
dis
 / p
suc
 = 1191.9 kPa / 497.59 kPa = 2.4 
 Figure 11-12 
 η
va
 = volumetric efficiency = 83 % 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
47.1354
100
83
L/s60w ×=
 
 w = 1.0565 kg/s at 0 C 
 
 P = w∆h
i
 
 ∆h
i
 = 21.5 kJ/kg 
 P = (1.0565)(21.5) 
 P = 22.7 kw at 0 C 
 
 
 (d) At 10 C evaporating temperature, Table A-6. 
 ν
suc
 = 34.7136 L/kg 
 p
suc
 = 680.70 kPa 
 Table A-7, 30 C 
 p
dis
 = 1191.9 kPa 
 
 Ratio = p
dis
 / p
suc
 = 1191.9 kPa / 680.70 kPa = 1.75 
 Figure 11-12 
 η
va
 = volumetric efficiency = 86.7 % 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
34.7136
100
86.7
L/s60w ×=
 
 w = 1.4986 kg/s at 10 C 
 
 P = w∆h
i
 
CHAPTER 11 - COMPRESSORS 
Page 4 of 6 
 ∆h
i
 = 13.7 kJ/kg 
 P = (1.4986)(13.7) 
 P = 20.5 kw at 10 C 
 
 (e) At 20 C evaporating temperature, Table A-6. 
 ν
suc
 = 26.0032 L/kg 
 p
suc
 = 909.93 kPa 
 Table A-7, 30 C 
 p
dis
 = 1191.9 kPa 
 
 Ratio = p
dis
 / p
suc
 = 1191.9 kPa / 909.93 kPa = 1.31 
 Figure 11-12 
 η
va
 =volumetric efficiency = 89.2 % 
 suc
vc
100ratentdisplacemew
ν
η
×=
 
 
( ) ( )
26.0032
100
89.2
L/s60w ×=
 
 w = 2.0583 kg/s at 20 C 
 
 P = w∆h
i
 
 ∆h
i
 = 6.5 kJ/kg 
 P = (2.0583)(6.5) 
 P = 13.4 at 20 C 
 
 
11-3. The catalog for a refrigerant 22, four-cylinder, hermetic compressor operating at 29 r/s. a condensing 
temperature of 40 C and an evaporating temperature of -4 C shows a refrigeration capacity of 115 kw. At this 
operating points the motor (whose efficiency is 90 percent) draws 34.5 kW. The bore of the cylinders is 87 
mm and the piston stroke is 70 mm. The performance data are based on 8C of subcooling of the liquid 
leaving the condenser. Compute (a) the actual volumetric efficiency and (b) the compression efficiency. 
 
Solution: Table A-6, -4 C evaporating temperature. 
 h
1
 = 403.876 kJ/kg 
 ν
suc
 = 53.5682 L/kg 
 s
1
 = 1.75775 kJ/kg.K 
 
 At 2, table A-7, constant entropy, 40 condensing temperature 
 h
2
 = 435.391 kJ/kg 
 ν
dis 
= 17.314 L/kg 
CHAPTER 11 - COMPRESSORS 
Page 5 of 6 
 At 3, 40 C condensing temperature, Table A-6, 8 C Subcooling 
 t = 40 -8 = 32 C 
 h
3
 = 239.23 kJ/kg 
 
 h
4
 = h
3
 = 239.23 kJ/kg 
 
 
 (a) For actual volumetric efficiency 
 Displacement rate = (4 cyl)(29 r/s)(0.087
2
pi / 4 m
3
/cyl.r)(0.070 m) 
 = 0.04827 m
3
/kg = 48.27 L/kg 
 
 Actual rate of refrigerant flow 
 = 115 kw / (403.876 - 239.23 kJ/kg) = 0.6985 kg/s 
 
 Actual volumetric flow rate at the compressor suction 
 = (0.6985 kg/s)(53.5682 L/kg) 
 = 37.42 L/s 
 
 
100
/smn,compressioofratentdisplaceme
/sm,compressorenteringrateflowvolume
3
3
va ×=η
 
 ηηηη
va
 = (37.42 L/s)(100) /(48.27 L/s) = 77.5 % - - - Ans. 
 
 (b) For compression efficiency. 
 Actual work of compression 
 = 0.9 (34.5 kW) / (0.6985 kg/s) = 44.45 kJ/kg 
 
100
kJ/kgn,compressioofworkactual
kJ/kgn,compressioof workisentropic
c ×=η
 
 
100
kJ/kg44.45
kJ/kg403.876-435.391
c ×=η
 
 ηηηη
c
 = 70.9 % - - - Ans. 
 
11-4. An automobile air conditioner using refrigerant 12 experiences a complete blockage of the airflow over the 
condenser, so that the condenser pressure rises until the volumetric efficiency drops to zero. Extrapolate the 
actual volumetric-efficiency curve of Fig. 11-12 to zero and estimate the maximum discharge pressure, 
assuming an evaporating temperature of 0 C. 
 
Solution: 
 
 Figure 11-12. 
 At actual volumetric efficiency = - 
 
 
( )
( ) ( ) 17.18576756
670
5ratioPressure =−
−
−
+=
 
 Table A-5, 0 C, p
suc
 = 308.61 kPa 
 
 p
dis
 = (17.18)(308.61 kPa) 
 p
dis
 = 5302 kPa - - - Ans. 
 
11-5. Compute the maximum displacement rate of a two-vane compressor having a cylinder diameter of 190 mm 
and a rotor 80 mm long with a diameter of 170 mm. The compressor operates at 29 r/s. 
 
CHAPTER 11 - COMPRESSORS 
Page 6 of 6 
Solution: Use Fig. 11-20 (a) 
 
 
 
θ = 3.3525 radians 
 
Crosshatched area = (1/2)(3.3525)(0.095)
2
 + (1/2)(0.094472)(0.010)(2) - (pi/2)(0.085)
2
 
 
Crosshatched area = 0.004724 m
2
. 
 
Displacement rate for two=vane compressor 
D = 2(Crosshatched area)(L)(rotative speed) 
D = (2)(0.004724)(0.080)(29) 
D = 0.0219 m
3
/s 
D = 21.9 L/s - - - Ans. 
 
11-6. A two-stage centrifugal compressor operating at 60 r/s is to compress refrigerant 11 from an evaporating 
temperature of 4 C to a condensing temperature of 35 C. If both wheels are to be of the same diameter, what 
is this diameter? 
 
Solution: 
 
 At 4 C evaporating temperature, Table A-4. 
 h
1
 = 390.93 kJ/kg 
 s
1
 = 1.68888 kJ/kg.K 
 
 At 35 C condensing temperature, Fig. A-2, constant entropy, 
 h
2
 = 410 kJ/kg 
 
 w = 60 r/s 
 Equation 11-16, 
 V
2t
2
 = 1000∆h
i
 
 V
2t
2
 = 1000(410 - 390.93)/2 
 V
2t
 = 97.65 m/s per stage 
 
 Section 11-25. Refrigerant 11. 113.1 m/s tip speed, 
 wheel diameter = 0.60 m 
 then at 97.65 m/s tip speed. 
 wheel diameter = (97.65 / 113.1)(0.6 m) 
 wheel diameter = 0.52 m - - - Ans. 
- 0 0 0 - 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 1 of 11 
12-1. An air-cooled condenser is to reject 70 kw of heat from a condensing refrigerant to air. The condenser has an 
air-side area of 210 m
2
 and a U value based on this area is 0.037 kW/m
2
.K; it is supplied with 6.6 m3/s of air, 
which has a density of 1.15 kg/m3. If the condensing temperature is to be limited to 55 C, what is the 
maximum allowable temperature of inlet air? 
 
Solution: A
o
 = 210 m
2
 
 U
o
 = 0.037 kW/m
2
.K 
 q = 70 kw 
 ρ = 1.15 kg/m3 
 Condensing Temperature = 55 C 
 w = (6.6 m
3
/s) / (1.15 kg/m
3
) = 5.739 kg/s 
 c
p
 =1.0 kJ/kg.K 
 
( ) ( )
( )( )



−
−
−−−
=
oc
ic
ocic
tt
tt
ln
tttt
LMTD
 
 q = U
o
A
o
LMTD 
 
 
( )( ) K9.0092100.037
70
AU
q
LMTD
oo
===
 
 But q = wc
p
(t
o
 - t
i
) 
 
( )( ) K12.197115.739
70
wc
q
tt
p
io ===−
 
 
( ) ( )
( )( )



−
−
−−−
=
oc
ic
ocic
tt
tt
ln
tttt
LMTD
 
 
( )( )



−
−
=
o
i
t55
t55
ln
12.197
9.009
 
 
3.8724
t55
t55
o
i
=
−
−
 
 55 - t
i
 = 3.8724(55 - 12.197 - t
i
) 
 t
i
 = 38.6 C - - - Ans. 
 
12-2. An air-cooled condenser has an expected U value of 30 W/m
2
.K based on the air-side area. The condenser 
is to transfer 60 kW with an airflow rate of 15 kg/s entering at 35 C. If the condenser temperature is to be 48 
C, what is the required air-side area? 
 
Solution: 
 q = U
o
A
o
LMTD 
 q = wc
p
(t
o
 - t
i
) 
 
 w = 15 kg/s 
 c
p
 = 1.0 kJ/kg.K 
 p
io
wc
q
tt +=
 
 ( )( )115
60
35t o +=
 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 2 of 11 
 t
o
 = 39 C 
 
( )
( )( )



−
−
−
=
oc
ic
io
tt
tt
ln
tt
LMTD
 
 
 
( )
( )( )
K10.878
3948
3548ln
3539
LMTD =




−
−
−
=
 
 q = U
o
A
o
LMTD 
 60 kw = (30 / 1000)(A
o
)(10.878) 
 A
o
 = 184 m
2
 - - - Ans. 
 
12-3. A refrigerant 22 condenser has four water passes and a total of 60 copper tubes that are 14 mm ID and have 
2 mm wall thickness. The conductivity of copper is 390 W/m.K. The outside of the tubes is finned so that the 
ratio of outside to inside area is 1.7. The cooling-water flow through the condenser tubes is 3.8 L/s. 
(a) Calculate the water-side coefficient if the water us at an average temperature of 30 C, at which 
temperature k = 0.614 W/m.K, ρ = 996 kg/m3, and m = 0.000803 Pa.s. 
(b) Using a mean condensing coefficient of 1420 W/m
2
.K, calculate the overall heat-transfer coefficient 
base don the condensing area. 
 
Solution: 
 
 (a) Water-side coefficient: 
 Eq. 12-19. 
 
0.4
p
0.8
k
cVD
0.023
k
hD







 µ






µ
ρ
=
 
 D = 14 mm = 0.014 m 
 k = 0.614 W/m.K 
 ρ = 996 kg/m3 
 µ = 0.000803 Pa.s 
 c
p
 = 4190 J/kg.K 
 
( )2
33
m0.014
44
60
/sm103.8
V





 pi






×
=
−
 
 V = 1.6457 m/s 
 
 
( ) ( )( )( ) ( )( ) 0.40.8
0.614
0.008034190
0.000803
9960.0141.6457
0.023
0.614
0.014h












=
 
 
 h = 7,313 W/m
2
.K - - - Ans. 
 
 (b) Overall heat-transfer coefficient. 
 Eq. 12-8. 
 iimoooo
Ah
1
kA
x
Ah
1
AU
1
++=
 
 ii
o
m
o
oo Ah
A
kA
xA
h
1
U
1
++=CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 3 of 11 
 h
o
 = 1420 W/m
2
.K 
 k = 390 W/m.K 
 A
o
 / A
i
 = 1.7 
 
( )io21m AAA += 
 






+=
1.7
A
AA oo2
1
m
 
 A
o
 / A
m
 = 1.25926 
 x = 2 mm = 0.002 m 
 h
i
 = 7,313 W/m
2
.K 
 
 
( )( )
7313
1.7
390
1.25960.002
1420
1
U
1
o
++=
 
 U
o
 = 1060 W/m
2
.K - - - Ans. 
 
12-4. A shell-and-tube condenser has a U value of 800 W/m
2
.K based on the water-side are and a water pressure 
drop of 50 kPa. Under this operating condition 40 percent of the heat-transfer resistance is on the water side. 
If the water-flow rate is doubled, what will the new U value and the new pressure drop be? 
 
Solution: 
 
 U
1
 = 800 W/m2.K 
 h
1
 = Water-side coefficient 
 
( )
2,000
800
1
0.40
1
h1 =






=
 
 Eq. 12-13, replace 0.6 by 0.8 for condenser. 
 Water-side coefficient = (const)(flow rate)
0.8
 
 
 For w
2
 / w
1
 = 2 
 
 
0.8
1
2
1
2
w
w
h
h






=
 
 h
2
 = (2000)(2)
0.8
 = 3482.2 W/m
2
.K 
 
 Remaining resistance = (0.60)( 1 / 800 ) = 0.00075 
 
 New U-Value: 
 
 
0.00075
3482.2
1
U
1
2
+=
 
 U
2
 = 964 W/m
2
.K - - - Ans. 
 
 New Pressure Drop: 
 
 Eq. 12-11. 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 4 of 11 
 
2
1
2
12
w
w
pp 





∆=∆
 
 
( )( )22 250p =∆ 
 ∆∆∆∆p
2
 = 200 kPa - - - Ans. 
 
12-5. (a) Compute the fin effectiveness of a bar fin made of aluminum that is 0.12 mm thick and 20 mm long when 
h
f
 = 28 W/m
2
.K, the base temperature is 4 C, and the air temperature is 20 C. 
 (b) If you are permitted to use twice as much metal for the fin as originally specified in part (a) and you can 
either double the thickness or double the length, which choice would be preferable in order to transfer the 
highest rate of heat flow. Why? 
 
Solution: 
 
 (a) Aluminum fins 
 k = 202 W/m.K 
 2y = 0.12 mm = 0.00012 m 
 y = 0.00006 m 
 L = 20 mm = 0.020 m 
 
ky
h
M f=
 
 
( )( )0.00006202
28
M =
 
 M = 48.1 m
-1
 
 
 ML
tanhML
=η
 
 ML = (48.1 m
-1
)(0.020 m) = 0.962 
 
( )
0.962
0.962tanh
=η
 
 ηηηη = 0.7746 - - - - Ans. 
 
 (b) If the fin thickness is doubled. 
 
 2y = 0.24 m = 0.00024 m 
 y = 0.00012 m 
 
( )( )0.00012202
28
M =
 
 M = 33.99 m
-1
 
 
 ML
tanhML
=η
 
 ML = (33.99 m
-1
)(0.020 m) = 0.6798 
 
( )
0.6798
0.6798tanh
=η
 
 η = 0.87 > 0.7746 
 If the length L is doubled 
 L = 40 mm = 0.040 m 
 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 5 of 11 
 
( )( )0.00006202
28
M =
 
 M = 48.1 m
-1
 
 
 ML
tanhML
=η
 
 ML = (48.1 m
-1
)(0.040 m) = 1.924 
 
( )
1.924
1.924tanh
=η
 
 η = 0.498 < 0.7746 
 
 
Ans. Therefore double the fin thickness to improve rate of heat flow with an efficiency of 87 % compared 
to 77.46 %. 
 
12-6. Compute the fin effectiveness of an aluminum rectangular plate fin of a finned air-cooling evaporator if the 
fins are 0.18 mm thick and mounted on a 16-mm-OD tubes. The tube spacing is 40 mm in the direction of air 
flow and 45 mm vertically. The air-side coefficient is 55 W/m
2
.K. 
 
Solution: h
f
 = 55 W/m
2
.K 
 
 Alumimum Fins, k = 202 W/m.K 
 2y = 0.00018 mm 
 y = 0.00009 mm 
 
ky
h
M f=
 
 
( )( )0.00009202
55
M =
 
 M = 55 m
-1
. 
 
 Equivalent external radius. 
 
( ) ( )( )
22
2
e
2
16
4540
2
16
r 





pi−=














−pi
 
 r
e
 = 23.94 mm = 0.02394 m 
 r
i
 = 8 mm = 0.008 m 
 (r
e
 - r
i
)M = (0.02394 - 0.008)(55) - 0.88 
 r
e
/r
i
 = 23.94 mm / 8 mm = 3 
 From Fig. 12-8/ 
 Fin Effectiveness = 0.68 - - - Ans. 
 
12-7. What is the UA value of a direct-expansion finned coil evaporator having the following areas: refrigerant side, 
15 m
2
; air-side prime, 13.5 m
2
, and air-side extended, 144 m
2
? The refrigerant-side heat-transfer coefficient 
is 1300 W/m
2
.K, and the air-side coefficient is 48 W/m
2
.K. The fin effectiveness is 0.64. 
 
Solution: η = 0.64 
 A
i
 = 15 m
2
 
 h
i
 = 1300 W/m
2
.K 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 6 of 11 
 h
f
 = 48 W/m
2
.K 
 A
p
 = 13.5 m
2
 
 A
e
 = 144 m
2
 
 
 Eq. 12-20 neglect tube resistance. 
 
( ) iiepfoo Ah
1
AAh
1
AU
1
+
η+
=
 
 
( ) ( )( ) ( )( )151300
11
AU
1
oo
+
+
=
14464.05.1348
 
 U
o
A
o
 = 4,025 W/K - - - Ans. 
 
12-8. A refrigerant 22 system having a refrigerating capacity of 55 kW operates with an evaporating temperature of 
5 C and rejects heat to a water-cooled condenser. The compressor is hermetically sealed. The condenser 
has a U value of 450 W/m
2
.K and a heat-transfer area of 18 m
2
 and receives a flow rate of cooling water of 
3.2 kg/s at a temperature of 30 C. What is the condensing temperature? 
 
Solution: Eq. 12-26. 
 
( ) ( )
( )( )



−
−
−−−
=
oc
ic
ocic
tt
tt
ln
tttt
LMTD
 
 
 Heat Rejection: 
 q = UALMTD = wc
p
(t
o
 - t
i
) 
 c
p
 = 4190 J/kg.K 
 
( )( ) ( )( )( )
( )( )( )30t41903.2
tt
t
ln
t
18450q o
oc
c
o
−=
















−
−
−
=
30
30
 
 
( )( ) 0.60412tttln occ =



−
− 30
 
 t
c
 - 30 = 1.82964 (t
c
 - t
o
) 
 t
o
 = 16.397 + 0.45345 t
c
 - - - Eq. No. 1 
 
 Figure 12-12. 
 At Heat-rejection ratio = 1.2 
 Condensing Temperature = 36 C 
 At Heat-rejection ratio = 1.3 
 Condensing Temperature = 49 C 
 
 Heat-rejection ratio = 0.92308 + 0.0076923 t
c
 
 
 
( )( ) ( )iopc ttwc550000.0076923t0.92308q −=+= 
 
( )( ) ( )( )( )30t41903.2550000.0076923t0.92308 oc −=+ 
 coc
0.45345t16.397t0.031554t33.7865 +==+
 
 t
c
 = 41.22 C - - - Ans. 
 
12-9. Calculate the mean condensing heat-transfer coefficient when refrigerant 12 condenses on the outside of the 
horizontal tubes in a shell-and-tube condenser. The outside diameter of the tubes is 19 mm, and in the 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 7 of 11 
vertical rows of tubes there are respectively, two, three, four, three, and two tubes. The refrigerant is 
condensing at a temperature of 52 C and the temperature of the tubes is 44 C. 
 
Solution: 
 Condensing Coefficient: Eq. 12-24. 
 
1/4
3
fg
2
cond
tND
khg
0.725h








∆µ
ρ
=
 
 Table A-5 at 52 C. 
 h
fg
 = 370.997 - 251.004 kJ/kg = 119.993 kJ/kg 
 h
fg
 = 199,993 J/kg 
 
 ρ = 1 / (0.83179 L/kg) = 1202 kg/m3 
 
 Table 15-5, Liquid Refrigerant 12 
 µ = 0.000179 PA.s 
 k = 0.05932 W/m.K 
 N = (2 + 3 + 4 + 3 +2) / 5 = 2.8 
 ∆t = 52 C - 44 C = 8 K 
 g = 9.81 m/s
2
 
 D = 19 mm = 0.019 m 
 
( )( ) ( )( )
( )( )( )( )
1/4
32
cond
0.0192.880.000174
0.05932119,99312029.81
0.725h








=
 
 h
cond
 = 1065 W/m
2
.K - - - Ans. 
 
12-10. A condenser manufacturer quarantees the U value under operating conditions to be 990 W/m
2
.K based on 
the water-side area. In order to allow for fouling of the tubes, what is the U value required when the 
condenser leaves the factory? 
 
Solution: 
 
 iff
o
o1o2 Ah
A
U
1
U
1
−=
 
 U
o1
 = 900 W/m
2
.K 
 1/ h
ff
 = 0.000176m
2
.K/W 
 A
o
 / A
i
 ~ 1.0 
 
( )10.000176
900
1
U
1
o2
−=
 
 U
o2
 = 1,199 W/m2.K - - - Ans. 
 
12-11. In example 12-3 the temperature difference between the refrigerant vapor and tube was originally assumed 
to be 5 K in order to compute the condensing coefficient. Check the validity of this assumption. 
 
Solution: 
 
( )( ) ( )( )
( ) ( )( )
4
1
0.0163.23t0.000180
0.0779160,90011099.81
0.725h
32
cond








∆
=
 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 8 of 11 
 
0.25cond t
2285
h
∆
=
 
 Then, 
 
( )
6910
1
0.014
0.016
0.000176
0.014
0.016
50.00000273
2285
t
U
1 0.25
o
⋅+++
∆
=
 
 
0.00036927
2285
t
U
1 0.25
o
+
∆
=
 
 
0.843782t
2285
U
0.25o +∆
=
 
 LMTD = 12.33 C 
 
 But 
 h
cond
∆t = U
o
LMTD 
 
( )12.33
0.843782t
2285
t
t
2285
0.250.25






+∆
=∆





∆ 
 0.843782t
12.33
t
0.25
0.75
+∆
=∆
 
 12.33t0.843782t
0.75
=∆+∆ 
 ∆t = 8.23 K 
 ∆t max= LMTD = 12.33 K 
 ∆t = 8.23 K to 12.33 K . . . Ans. 
 
12-12. (a) A Wilson plot is to be constructed for a finned air-cooled condenser by varying the rate of airflow. What 
should the abscissa of the plot be? 
 (b) A Wilson plot is to be constructed for a shell-and-tube water chiller in which refrigerant evaporates in 
tubes. The rate of water flow is to be varied for the Wilson plot. What should the abscissa of the plot be? 
 
Solution: 
 (a) Eq. 12-20. 
 
 
( ) iimepfoo Ah
1
kA
x
AAh
1
AU
1
++
η+
=
 
 Eq. 12-21 
 
0.5
f 38Vh = 
 V in m/s. 
 Varying airflow, the Wilson plot is a graph of 1/U
o
 versus 1/V
0.5
. 
 Abscissa is 1/V
0.5
 where V is the face velocity in meters per second. 
 
 (b) Eq. 12-27 
 iimooo
Ah
1
kA
x
h
1
AU
1
++=
 
 Liquid in shell, variation of Eq. 12-13, 
 
( ) 0.6o Vconsth = 
 Varying water flow, the Wilson plot is a graph of 1/U
o
 versus 1/V
0.6
. 
 Abscissa is 1/V
0.6
 where V is the face velocity in meters per second. 
 
12-13. The following values were measured on an ammonia condenser. 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 9 of 11 
 
Uo, W/sq m.K 2300 2070 1930 1760 1570 1360 1130 865 
V, m/s 1.22 0.975 0.853 0.731 0.61 0.488 0.366 0.244 
 
 Water flowed inside the tubes, and the tubes were 51 mm OD and 46 mm ID and had a conductivity of 60 
W/m.K. Using a Wilson plot, determine the condensing coefficient. 
 
Solution: 
 Wilson plot 
 
0.8
2
1
o V
C
C
U
1
+=
 
 Tabulation: 
 
 1/U
o
 1/V
0.8
 
 
0.000434783 0.852928 
0.000483092 1.020461 
0.000518135 1.13564 
0.000568182 1.28489 
0.000636943 1.485033 
0.000735294 1.775269 
0.000884956 2.234679 
0.001156069 3.090923 
 
 By linear regression: 
 C
1
 = 0.000153033 
 C
2
 = 0.000325563 
 
 But: 
 m
o
o
1
kA
xA
h
1
C +=
 
 
( ) 1.0515524651
51
A
A
m
o
=
+
=
 
 x = (1/2)(51 - 46) = 2.5 mm = 0.0025 m 
 k = 60 W/m.K 
 
( )( )
60
1.051550.0025
h
1
30.00015303
o
+=
 
 h
o
 = 9,156 W/m
2
.K - - - Ans. 
 
12-14. Develop Eq. (12-23) from Eq. (12-22). 
 
Solution: Eq. 12-22. 
 
4
1
3
fg
2
cv
tk4
xhg
k
xh








∆µ
ρ
=
 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 10 of 11 
 L
dxh
h
L
0
cv
cv
∫
=
 
 L
dx
tk4
xhg
x
k
h
L
0
4
1
3
fg
2
cv
∫ 







∆µ
ρ
=
 
 
∫







∆µ
ρ






=
L
0
4
1-
4
1
fg
2
cv dxx
tk4
hg
L
k
h
 
 
L
0
4
3
4
1
fg
2
cv x
3
4
tk4
hg
L
k
h 

















∆µ
ρ






=
 
 
4
3
4
1
fg
2
cv L
3
4
tk4
hg
L
k
h 













∆µ
ρ






=
 
 
4
1
3
fg
2
4
1
cv
tLk
khg
3
4
4
1
h








∆µ
ρ












=
 
 Ans. Eq. 12-23. 
 
4
1
3
fg
2
cv
tLk
khg
0.943h








∆µ
ρ
=
 
 
12-15. From Fig. 12-21, determine C and b in the equation h = C∆Tb applicable to values in the middle of the typical 
range. 
 
 
Solution: Use Fig. 12-21 
 
Tabulation: 
Heat-transfer Coefficient Heat flux ∆t 
 W/m
2
.K, h W/m
2
 K 
 
 400 710 1.775 
 600 1550 2.583 
 800 2820 3.525 
 1000 4170 4.170 
 1500 9000 6.000 
 
 h = C∆tb 
 
 By Curve-Fitting: 
 C = 212.8 
 b = 1.08 
 
 h = 212.8∆∆∆∆t1.08 - - - Ans. 
 
13-16. Section 12-18 makes the statement that on a graph of the performance of a water chilling evaporator with the 
coordinates of Fig. 12-23, a curve for a given entering water temperature is a straight line if the heat-transfer 
CHAPTER 12 - CONDENSERS AND EVAPORATORS 
 Page 11 of 11 
coefficients are constant. prove this statement. 
 
Solution: Use Fig. 12-23. 
 
 t
e
 = evaporating temperature 
 t
a
 = entering-water temperature (constant) 
 U = heat-transfer coefficient (constant) 
 
 
( ) 





−
+
=−= e
ba
abp t
2
tt
UAttwcq
 
 ebaapbp
UAt0.5UAt0.5UAttwctwc −+=−
 
 
( )
( )0.5UAwc
UAtt0.5UAwc
t
p
eap
b
−
−+
=
 
 
( )
( ) 





−
−
−+
= a
p
eap
p t
0.5UAwc
UAtt0.5UAwc
wcq
 
 
( ) ( )
( ) 




−
−−−+
=
0.5UAwcp
t0.5UAwcUAtt0.5UAwc
wcq
apeap
p
 
 
( )





−
−
=
0.5UAwc
UAtUAt
wcq
p
ea
p
 
 
( )ea
p
p
tt
0.5UAwc
UAwc
q −
−
=
 
 If U is constant. 
 q = (constant)(t
a
 - t
e
) 
 At constant t
0
, this is a straight line. - - - Ans. 
 
- 0 0 0 - 
CHAPTER 13 - EXPANSION DEVICES 
Page 1 of 14 
13-1. Using the method described in Sec. 13-5 and entering conditions given in Table 13-1 for example 13-1 at 
position 4, compute the length of tube needed to drop the temperature to 36 C. Use property values from 
Refrigerant 22 tables when possible. 
 
Solution: 
 At Table 13-1, position 4 
 Temperature = 37 C. 
 p
4
 = 1425.8 kPa 
 x
4
 = 0.023 
 ν
4
 = 0.001230 m
3
/kg 
 h
4
 = 249.84 kJ/kg 
 V
4
 = 5.895 m/s 
 
 At Position 5, t = 36 C 
 Eq. 13-15 
 
273.15t
2418.4
15.06
1000
p
ln
+
−=





 
 
273.1536
2418.4
15.06
1000
p
ln 5
+
−=





 
 p
5
 = 1390.3 kPa 
 Eq. 13-16. 
 
2
f5 t0.000016080.002062t0.777 ++=ν 
 
( ) ( )
1000
360.00001608360.0020620.777
2
f5
++
=ν
 
 n
f5
 = 0.000872 m
3
/kg 
 Eq. 13-17. 
 
( )
1000
p273.15t940504.26
g5
++−
=ν
 
 
( ) ( )
1000
1390300273.1536940504.26
g5
++−
=ν
 
 n
f5
 = 0.01665 m
3
/kg 
 Eq. 13-18. 
 
2
f5 0.001854t1.172t200.0 ++=h 
 
( ) ( )2f5 360.001854361.172200.0h ++= 
 h
f5
 = 244.6 kJ/kg 
 Eq. 13-19 
 
2
g5 0.002273t0.3636t405.5h −+= 
 
( ) ( )2g5 360.002273360.3636405.5h −+= 
 h
g5
 = 415.64 kJ/kg 
 Eq. 13-20 
 
296
f5 t108.869t101.7150.0002367
−− ×+×−=µ
 
 
( ) ( )296f5 36108.86936101.7150.0002367 −− ×+×−=µ 
 µ
f5
 = 0.0001865 Pa.s 
 Eq. 13-21.2996
g5 t100.2560t1050.061011.945
−−− ×+×+×=µ
 
CHAPTER 13 - EXPANSION DEVICES 
Page 2 of 14 
 
( ) ( )2996g5 36100.2560361050.061011.945 −−− ×+×+×=µ 
 µ
g5
 = 0.00001408 Pa.s 
 Eq. 13-14. 
 2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f5g5 





ν−ν=
 
 w/A = 4792.2 kg/s.m
2
 from Ex. 13-1. 
 
( ) ( ) 2858.54
2
1
4792.20.0008720.01665a
22
=−=
 
 
( ) ( ) 2f5g5f5f5g5
A
w
hh1000b 





ν−νν+−=
 
 
( ) ( )( )24792.20.0008720.016650.00872244.6415.641000b −+−=
 
 b = 171,356 
 
( )
2
V
2
1
A
w
hh1000c
2
4
f5
2
4f5 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
5.895
0.000872
2
1
4792.2249.84244.61000c
2
22
−+−=
 
 c = -5,248.65 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.0312858.542
5248.65-2854.544171,356171,356
x
2
=
−±−
=
 
 Then: 
 h
5
 = h
f5
 + x(h
g5
 - h
f5
) 
 h
5
 = 244.6 + 0.031 (415.64 - 244.6) 
 h
5
 = 249.9 kJ/kg 
 ν
5
 = ν
f5
 + x(ν
g5
 - ν
f5
) 
 ν
5
 = 0.000873 + 0.031 (0.01665 - 0.000872) 
 ν
5
 = 0.001361 m
3
/kg 
 
( )f5g5f55 x µ−µ+µ=µ 
 
( )0.00018650.000014080.0310.00018655 −+=µ 
 µ
5
 = 0.0001812 Pa.s 
 
( )( )0.0013614792.2
A
w
V 55 =ν=
 
 V
5
 = 6.522 m/s 
 






µ
=
µν
=
A
wDVD
Re
 
 D = 1.63 mm = 0.00163 m 
 At 4: 
 
296
f44 t108.869t101.7150.0002367
−− ×+×−=µ=µ
 
 t
4
 = 37 C 
CHAPTER 13 - EXPANSION DEVICES 
Page 3 of 14 
 
( ) ( )2964 37108.86937101.7150.0002367 −− ×+×−=µ 
 
0.00018544 =µ 
 
( )
( ) ( ) 43,1094792.20.0001812
0.00163
Re ==
 
 Eq. 13-9. 
 
0.25Re
0.33
f =
 
 ( )
0.02303
42,132
0.33
f
0.254
==
 
 ( )
0.02290
43,109
0.33
f
0.255
==
 
 
0.022965
2
0.022900.02303
fm =
+
=
 
 
m/s6.2085
2
6.5225.895
Vm =
+
=
 
 Eq. 13-4 
 
( ) ( )45
2
54 VVwA
2
V
D
L
f-p-p −=








ν
∆
 
 Eq. 13-7 
 A
w
2
V
D
L
f
2
V
D
L
f
2 ∆
=
ν
∆
 
 
( ) ( )4554 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
 
( ) ( )5.8956.5224792.2
A
w
2
V
D
L
f-1390.3-1425.81000 −=
∆
 
 
32,495.3
A
w
2
V
D
L
f =
∆
 
 
( ) ( )
( ) ( ) 32,495.34792.2
2
6.2085
0.00163
L
0.022965 =
∆
 
 ∆∆∆∆L
4-5
 = 0.155 m - - - Ans. 
 
13-2. A capillary tube is to be selected to throttle 0.011 kg/s of refrigerant 12 from a condensing pressure of 960 
kPa and a temperature of 35 C to an evaporator operating at -20 C. 
(a) Using Figs. 13-7 and 13-8, select the bore and length of a capillary tube for this assignment. 
(b) If the evaporating temperature had been 5 C rather than -20 C, would the selection of part (a) be 
suitable? Discuss assumptions that have been made. 
 
Solution: Table A-5, p = 960 kPa, t
sat
 = 40 C, 
 Subcooling = 40 C - 35 C = 5 C 
 (a) Use bore diameter D = 1.63 mm 
 Fig. 13-7, 960 kPa inlet pressure, saturated. 
 Flow rate = 0.0089 kg/s 
 Fig. 13-8. 
 Flow correction factor = (0.011 kg/s)/(0.0089 kg/s) 
 Flow correction factor = 1.24 
 Then Length = 1,230 mm = 1.23 m L - - - Ans. 
 (b) Use positions from 35 C to -20 C at 5 C increment. 
 Table A-5, 35 C, sat. p = 847.72 kPa. 
CHAPTER 13 - EXPANSION DEVICES 
Page 4 of 14 
 At position 1, 
 h
1
 = 233.50 kJ/kg 
 ν
1
 = 0.78556 L/kg = 0.000786 m
3
/kg 
 Table 15-5, µ
1
 = 0.000202 Pa.s 
 p
1
 = 960 kPa 
 ( )
2
2
kg/s.m5271.4
40.00163
0.011
A
w
=
pi
=
 
 
( )( )0.0007865271.4
A
w
V 11 =ν=
 
 V
1
 = 4.143 m/s 
 






µ






=
νµ
=
111
1
1
D
A
wDV
Re
 
 
( )( )
42,537
0.000202
0.001635271.4
Re1 ==
 
 
( ) 0.0229842537
0.33
Re
0.33
f
0.250.25
1
1 ===
 
 At position 2, 30 C 
 p
2
 = 744.90 kPa 
 h
f2
 = 228.54 kJ/kg 
 h
g2
 = 363.57 kJ/kg 
 ν
f2
 = 0.77386 L/kg = 0.000774 m
3
/kg 
 ν
g2
 = 23.5082 L/kg = 0.02351 m
3
/kg 
 µ
f2
 = 0.0002095 Pa.s 
 µ
g2
 = 0.00001305 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f2g2 





ν−ν=
 
 
( ) ( ) 1.7182=−=
2
1
5271.40.0007740.02351a
22
 
 
( ) ( ) 2f2g2f2f2g2
A
w
hh1000b 





ν−νν+−=
 
 
( ) ( )( )25271.40.0007740.023510.000774228.54363.571000b −+−=
 
 b = 135,519 
 
( )
2
V
2
1
A
w
hh1000c
2
1
f2
2
1f2 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
4.143
0.000774
2
1
5271.4233.50228.541000c
2
22
−+−=
 
 c = -4,960.3 
 2a
4acbb
x
2
−±−
=
 
CHAPTER 13 - EXPANSION DEVICES 
Page 5 of 14 
 
( )( )
( ) 0.03657182.12
4960.3-7182.14135,519135,519
x
2
=
−±−
=
 
 Then: 
 h
2
 = h
f2
 + x(h
g2
 - h
f2
) 
 h
2
 = 233.47 kJ/kg 
 ν
2
 = ν
f2
 + x(ν
g2
 - ν
f2
) 
 ν
2
 = 0.001604 m
3
/kg 
 
( )f2g2f22 x µ−µ+µ=µ 
 
 µ
2
 = 0.0002023 Pa.s 
 
( )( )0.0016045271.4
A
w
V 22 =ν=
 
 V
2
 = 8.455 m/s 
 






µ
=
µν
=
A
wDVD
Re
2
2
 
 
( )( )
42,474
0.0002023
0.001635271.4
Re 2 ==
 
 
0.252 Re
0.33
f =
 
 ( )
0.02299
42,474
0.33
f
0.252
==
 
 
 
0.022985
2
0.022990.02298
fm =
+
=
 
 
m/s
2
4.142
Vm 299.6
455.8
=
+
=
 
 
( ) ( )1221 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )4.1438.4555271.45271.4
2
6.299
0.00163
L
0.022985-744.9-9601000 −=
∆
 
 ∆L
1-2
 = 0.8217 m 
 
 At position 3, 25 C 
 p
2
 = 651.62 kPa 
 h
f2
 = 223.65 kJ/kg 
 h
g2
 = 361.68 kJ/kg 
 ν
f2
 = 0.76286 L/kg = 0.000763 m
3
/kg 
 ν
g2
 = 26.8542 L/kg = 0.026854 m
3
/kg 
 µ
f2
 = 0.000217 Pa.s 
 µ
g2
 = 0.0000128 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f3g3 





ν−ν=
 
CHAPTER 13 - EXPANSION DEVICES 
Page 6 of 14 
 
( ) ( ) 1.9458=−=
2
1
5271.40.0007630.026854a
22
 
 
( ) ( ) 2f3g3f3f3g3
A
w
hh1000b 





ν−νν+−=
 
 
( ) ( )( )25271.40.0007630.0268540.000763223.65361.681000b −+−=
 
 b = 138,583 
 
( )
2
V
2
1
A
w
hh1000c
2
2
f3
2
2f3 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
8.455
0.000763
2
1
5271.4233.47223.651000c
2
22
−+−=
 
 c = -9,847.7 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.07079458.12
9847.7-9458.14138,583138,583
x
2
=
−±−
=
 
 Then: 
 h
3
 = h
f3
 + x(h
g3
 - h
f3
) 
 h
3
 = 233.41 kJ/kg 
 ν
3
 = ν
f3
 + x(ν
g3
 - ν
f3
) 
 ν
3
 = 0.002608 m
3
/kg 
 
( )f3g3f33 x µ−µ+µ=µ 
 
 µ
3
 = 0.0002026 Pa.s 
 
( )( )0.0026085271.4
A
w
V 33 =ν=
 
 V
3
 = 13.748 m/s 
 






µ
=
µν
=
A
wDVD
Re
3
3
 
 
( )( )
42,411
0.0002026
0.001635271.4
Re 3 ==
 
 
0.253 Re
0.33
f =
 
 ( )
0.0230
42,411
0.33
f
0.253
==
 
 
 
0.0230
2
0.02300.02299
fm =
+
=
 
 
m/s
2
8.455
Vm 102.11
748.13
=
+
=
 
 
( ) ( )2332 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )8.45513.7485271.45271.4
2
11.102
0.00163
L0.0230-651.62-744.91000 −=
∆
∆L
2-3
 = 0.1584 m 
 
CHAPTER 13 - EXPANSION DEVICES 
Page 7 of 14 
 At position 4, 20 C 
 p
4
 = 567.29 kPa 
 h
f4
 = 218.82 kJ/kg 
 h
g4
 = 359.73 kJ/kg 
 ν
f4
 = 0.75246 L/kg = 0.00075246 m
3
/kg 
 ν
g4
 = 30.7802 L/kg = 0.0307802 m
3
/kg 
 µ
f2
 = 0.000225 Pa.s 
 µ
g2
 = 0.0000126 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f4g4 





ν−ν=
 
 
( ) ( ) 528,12=−=
2
1
5271.40.000752460.0307802a
22
 
 
( ) ( ) 2f4g4f4f4g4
A
w
hh1000b 





ν−νν+−=
 
( ) ( )( )25271.40.000752460.03078020.00075246218.82359.731000b −+−=
 
 b = 141,538 
 
( )
2
V
2
1
A
w
hh1000c
2
3
f4
2
3f4 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
13.748
0.00075246
2
1
5271.4233.41218.821000c
2
22
−+−=
 
 c = -14,677 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.1028125282
14677-125284141,538141,538
x
2
=
−±−
=
 
 Then: 
 h
4
 = h
f4
 + x(h
g4
 - h
f4
) 
 h
4
 = 233.31 kJ/kg 
 ν
4
 = ν
f4
 + x(ν
g4
 - ν
f4
) 
 ν
4
 = 0.003839 m
3
/kg 
 
( )f4g4f44 x µ−µ+µ=µ 
 
 µ
4
 = 0.0002032 Pa.s 
 
( )( )0.0038395271.4
A
w
V 44 =ν=
 
 V
4
 = 20.237 m/s 
 






µ
=
µν
=
A
wDVD
Re
4
4
 
 
( )( )
42,285
0.0002032
0.001635271.4
Re 4 ==
 
CHAPTER 13 - EXPANSION DEVICES 
Page 8 of 14 
 
0.254 Re
0.33
f =
 
 ( )
0.0230
42,285
0.33
f
0.254
==
 
 
 
0.0230
2
0.02300.0230
fm =
+
=
 
 
m/s
2
13.748
Vm 993.16
237.20
=
+
=
 
 
( ) ( )3434 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )13.74820.2375271.45271.4
2
16.993
0.00163
L
0.0230-567.29-651.621000 −=
∆
∆L
3-4
 = 0.0793 m 
 
 At position 5, 15 C 
 p
5
 = 491.37 kPa 
 h
f5
 = 214.05 kJ/kg 
 h
g5
 = 357.73 kJ/kg 
 ν
f5
 = 0.74262 L/kg = 0.00074262 m
3
/kg 
 ν
g5
 = 35.4133 L/kg = 0.0354133 m
3
/kg 
 µ
f5
 = 0.0002355 Pa.s 
 µ
g5
 = 0.0000124 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f5g5 





ν−ν=
 
 
( ) ( ) 16,701
2
1
5271.40.000742620.0354133a
22
=−=
 
 
 
( ) ( ) 2f5g5f5f5g5
A
w
hh1000b 





ν−νν+−=
 
( ) ( )( )25271.40.000742620.03541330.00074262214.05357.731000b −+−=
 
 b = 144,396 
 
( )
2
V
2
1
A
w
hh1000c
2
4
f5
2
4f5 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
20.237
0.00074262
2
1
5271.4233.31214.051000c
2
22
−+−=
 
 c = -19,457 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.132716,7012
19,457-16,7014144,396144,396
x
2
=
−±−
=
 
 Then: 
 h
5
 = h
f5
 + x(h
g5
 - h
f5
) 
CHAPTER 13 - EXPANSION DEVICES 
Page 9 of 14 
 h
5
 = 233.12 kJ/kg 
 ν
5
 = ν
f5
 + x(ν
g5
 - ν
f5
) 
 ν
5
 = 0.005343 m
3
/kg 
 
( )f5g5f55 x µ−µ+µ=µ 
 
 µ
5
 = 0.0002059 Pa.s 
 
( )( )0.0053435271.4
A
w
V 55 =ν=
 
 V
5
 = 28.165 m/s 
 






µ
=
µν
=
A
wDVD
Re
5
5
 
 
( )( )
41,731
0.0002059
0.001635271.4
Re 5 ==
 
 
0.255 Re
0.33
f =
 
 ( )
0.02309
41,731
0.33
f
0.255
==
 
 
 
0.02305
2
0.023090.0230
fm =
+
=
 
 
m/s
2
20.237
Vm 201.24
165.28
=
+
=
 
 
( ) ( )4554 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )20.23728.1655271.45271.4
2
24.201
0.00163
L
0.02305-491.37-567.291000 −=
∆
∆L
4-5
 = 0.0378 m 
 
 At position 6, 10 C 
 p
6
 = 423.30 kPa 
 h
f6
 = 209.32 kJ/kg 
 h
g6
 = 355.69 kJ/kg 
 ν
f6
 = 0.73326 L/kg = 0.00073326 m
3
/kg 
 ν
g6
 = 40.9137 L/kg = 0.0409137 m
3
/kg 
 µ
f6
 = 0.000246 Pa.s 
 µ
g6
 = 0.0000122 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f6g6 





ν−ν=
 
 
( ) ( ) ,431
2
1
5271.40.000733260.0409137a
22 22=−=
 
 
 
( ) ( ) 2f6g6f6f6g6
A
w
hh1000b 





ν−νν+−=
 
CHAPTER 13 - EXPANSION DEVICES 
Page 10 of 14 
( ) ( )( )25271.40.000733260.04091370.00073326209.32355.691000b −+−=
 
 b = 147,189 
 
( )
2
V
2
1
A
w
hh1000c
2
5
f6
2
5f6 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
28.165
0.00073326
2
1
5271.4233.12209.321000c
2
22
−+−=
 
 c = -24,189 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.160422,4312
24,189-22,4314147,189147,189
x
2
=
−±−
=
 
 Then: 
 h
6
 = h
f6
 + x(h
g6
 - h
f6
) 
 h
6
 = 232.80 kJ/kg 
 ν
6
 = ν
f6
 + x(ν
g6
 - ν
f6
) 
 ν
6
 = 0.007178 m
3
/kg 
 
( )f6g6f66 x µ−µ+µ=µ 
 
 µ
6
 = 0.0002085 Pa.s 
 
( )( )0.0071785271.4
A
w
V 66 =ν=
 
 V
6
 = 37.838 m/s 
 






µ
=
µν
=
A
wDVD
Re
6
6
 
 
( )( )
41,211
0.0002085
0.001635271.4
Re 6 ==
 
 
0.256 Re
0.33
f =
 
 ( )
0.02316
41,211
0.33
f
0.256
==
 
 
 
0.02313
2
0.023160.02309
fm =
+
=
 
 
m/s
2
28.165
Vm 33
838.37
=
+
=
 
 
( ) ( )5665 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )28.16537.8385271.45271.4
2
33.0
0.00163
L
0.02313-423.30-491.371000 −=
∆
∆L
5-6
 = 0.0138 m 
 
 At position 7, 5 C 
 p
7
 = 363.55 kPa 
 h
f7
 = 204.64 kJ/kg 
 h
g7
 = 353.60 kJ/kg 
CHAPTER 13 - EXPANSION DEVICES 
Page 11 of 14 
 ν
f7
 = 0.72438 L/kg = 0.00072438 m
3
/kg 
 ν
g7
 = 47.4853 L/kg = 0.0474853 m
3
/kg 
 µ
f6
 = 0.0002565 Pa.s 
 µ
g6
 = 0.0000120 Pa.s 
 
2a
4acbb
x
2
−±−
=
 
 
( )
2
1
A
w
a
2
2
f7g7 





ν−ν=
 
 
( ) ( ) 30,380
2
1
5271.40.000724380.0474853a
22
=−=
 
 
 
( ) ( ) 2f7g7f7f7g7
A
w
hh1000b 





ν−νν+−=
 
( ) ( )( )25271.40.000724380.04748530.00072438204.64353.601000b −+−=
 
 b = 149,901 
 
( )
2
V
2
1
A
w
hh1000c
2
6
f7
2
6f7 −ν





+−= 2
 
 
( ) ( ) ( ) ( )
2
37.838
0.00072438
2
1
5271.4232.80204.641000c
2
22
−+−=
 
 c = -28,869 
 2a
4acbb
x
2
−±−
=
 
 
( )( )
( ) 0.185630,3802
28,869-30,3804149,901149,901
x
2
=
−±−
=
 
 Then: 
 h
7
 = h
f7
 + x(h
g7
 - h
f7
) 
 h
7
 = 232.29 kJ/kg 
 ν
7
 = ν
f7
 + x(ν
g7
 - ν
f7
) 
 ν
7
 = 0.009403 m
3
/kg 
 
( )f7g7f77 x µ−µ+µ=µ 
 
 µ
7
 = 0.0002111 Pa.s 
 
( )( )0.0094035271.4
A
w
V 77 =ν=
 
 V
6
 = 49.567 m/s 
 






µ
=
µν
=
A
wDVD
Re
7
7
 
 
( )( )
40,703
0.0002111
0.001635271.4
Re 7 ==
 
 
0.257 Re
0.33
f =
 
 ( )
0.02323
40,703
0.33
f
0.257
==
 
CHAPTER 13 - EXPANSION DEVICES 
Page 12 of 14 
 
 
0.02320
2
0.023230.02316
fm =
+
=
 
 
m/s
2
37.838
Vm 703.43
567.49
=
+
=
 
 
( ) ( )6776 VV
A
w
A
w
2
V
D
L
f-p-p −=
∆
 
( ) ( ) ( )
( ) ( ) ( )37.83849.5675271.45271.4
2
43.703
0.00163
L
0.02320-363.55-423.31000 −=
∆
∆L
6-7
 = -0.0013 m ~ 0.000m 
 
Assume choked flow is at approximately 5 C. 
 L = ∆L
1-2
 +∆L
2-3
 +∆L
3-4
 +∆L
4-5
 +∆L
5-6
 +∆L
6-7
 
 L = 0.8217 m + 0.1584 m + 0.0793 m + 0.0378 m + 0.0138 m + 0 m 
 L = 1.111 m 
 
 
Ans. By assuming choked flow length the same , choked flow is at 5 C. 5 C is still suitable for the selection 
of part (a) as it is the choked flow temperature. 
 
13-3. A refrigerant 22 refrigerating system operates with a condensing temperature of 35 C and an evaporating 
temperature of -10 C. If the vapor leaves the evaporator saturated and is compressed isentropically, what is 
the COP of the cycle (a) if saturated liquid enters the expansion device and (b) if the refrigerant entering the 
expansion device is 10 percent vapor as in Fig. 13-3? 
 
Solution: Table A-6. 
 At 1, -10 C, h
1
 = 401.555 kJ/kg 
 s
1
 = 1.76713 kJ/kg 
 
 At 2, 35 C, constant entropy, Table A-7 
 h
2
 = 435.212 kJ/kg 
 
 (a) At 35 C saturated. 
 h
3
 = h
f
 = 243.114 kJ/kg 
 h
4
 = h
3
 = 243.114 kJ/kg 
 
401.555435.212
243.114401.555
hh
hh
COP
12
41
−
−
=
−
−
=
 
 COP = 4.71 - - - Ans. 
 
 (b) h
3
 = h
f
 + x (h
g
 - h
f
) 
 h
f
 = 243.114 kJ/kg 
 h
g
 = 415.627 kJ/kg 
 x = 0.10 
 
 h
3
= 243.114 + (0.10)(415.627 - 243.114) 
 h
3
 = 260.365 kJ/kg 
 h
4
 = h
3
 = 260.365 kJ/kg 
 
 
401.555435.212
260.365401.555
hh
hh
COP
12
41
−
−
=
−
−
=
 
 COP = 4.20 - - - Ans. 
CHAPTER 13 - EXPANSION DEVICES 
Page 13 of 14 
 
13-4. Refrigerant 22 at a pressure of 1500 kPa leaves the condenser and rises vertically 10 m to the expansion 
valve. The pressure drop due to friction in the liquid line is 20 kPa. In order to have no vapor in the refrigerant 
entering the expansion valve, what is the maximum allowable temperature at that point? 
 
Solution: say ν = ν
1
 
 p
2
 = p
1
 - gH/ν
1
 - ∆p 
 H = 10 m 
 g = 9.81 m/s 
 ∆p = 20 kPa 
 p
1
 = 1500 kPa 
 Table A-6. 
 ν
1
=0.8808 L/kg = 0.0008808 m
3
/kg 
 p
2
 = (1500)(1000) - (9.81)(10) / (0.0008808) - (20)(1000) 
 p
2
 = 1,368.62 kPa 
 Table A-6. 
 t
2
 = 35.4 C - - - Ans. 
 
13-5. A superheat-controlled expansion valve in a refrigerant 22 system is not equipped with an external equalizer. 
The valve supplies refrigerant to an evaporator coil and comes from the factory with a setting that requires 
5K superheat in order to open the valve at an evaporator temperature of 0 C. 
(a) What difference in pressure on opposite sides of the diaphragm is required to open the valve? 
(b) When the pressure at the entrance of the evaporator is 600 kPa, how much superheat is required 
to open the valve if the pressure drop of the refrigerant through the coil is 55 kPa? 
 
Solution: 
 Using Fig. 13-15 and deriving equation by assuming parabolic curve. 
 Let y - pressure, kPa and x = temperature , C. 
 
 y
2
 - y
1
 = A (x
2
2
 - x
1
2
) + B(x
2
 - x
1
) 
 
 At 5 C evaporator temperature, 5 K superheat 
 100 kPa pressure differential 
 x
1
 = 5 C, x
2
 = 5 C + 5 = 10 C 
 y
2
 - y
1
 = 100 kPa 
 
 100 = A (10
2
 - 5
2
) + B (10 - 5) 
 100 = 75A + 5B - - Eq. 1 
 
 At -30 C evaporator temperature 
 12 C superheat 
 100 kPa pressure differential 
 x
1
 = -30 C, x
2
 = -30 C + 12 = -18 C 
 y
2
 - y
1
 = 100 kPa 
 
 100 = A ((-18)
2
 -(-30)
2
) + B ((-18) -(-30)) 
 100 = -576A + 12 B - - Eq. 2 
 
 But 5B = 100 - 75A 
 Then 100 = -576A + 12 (20 - 15A) 
 A = 0.185185 
 B = 17.222222 
CHAPTER 13 - EXPANSION DEVICES 
Page 14 of 14 
 
 Therefore: 
 y
2
 - y
1
 = 0.185185 (x
2
2
 - x
1
2
) + 17.222222(x
2
 - x
1
) 
 
 (a) At 0 C evaporator temperature, 5 K superheat 
 x
1
 = 0 C 
 x
2
 = 0 C + 5 = 5 C 
 y
2
 - y
1
 = 0.185185 (5
2
 -0
2
) + 17.222222(5 -0) 
 y
2
 - y
1
 = 90.74 kPa - - - Ans. 
 
 
 (b) At 0 C evaporator temperature, p = 497.59 kPa 
 ∆p = 600 kPa + 55 kPa - 497.59 kPa = 157.41 kPa. 
 x
1
 = 0 C 
 Then 
 157.41 = 0.185185 (x
2
2
 -0
2
) + 17.222222(x
2
 -0) 
 x
2
 = 8.4 C 
 x
2
 - x
1
 = 8.4 K - - - Ans. 
 
13-6. The catalog of an expansion valve manufacturer specifies a refrigerating capacity of 45 kW for a certain 
valve when the pressure difference across the valve is 500 kPa. The catalog ratings apply when vapor-free 
liquid at 37.8 C enters the expansion valve and the evaporator temperature is 4.4 C. What is the expected 
rating of the valve when the pressure difference across it is 1200 kPa? 
 
Solution: 
 Eq. 13-22 
 
( ) m/sfferencepressuredi2CVelocity =
 
 With all other data as constant except for pressure difference and refrigerating capacity. 
 
( ) m/sfferencepressuredi2Capacity ingRefrigerat α
 
 
 Then: 
 New Refrigerating Capacity 
 
( )
kPa500
kPa1200
kW45=
 
 = 69.7 kW - - - - Ans. 
 
- 0 0 0 - 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 1 of 11 
14-1. Either graphically or by using the computer, for an ambient temperature of 30 C develop the performance 
characteristics of a condensing unit (of the form of Fig. 14-6 or Table 14-3) if the compressor has 
performance shown by Fig. 14-1 [ or Eq. (14-1) and (14-2)] and the condenser has characteristics shown by 
Fig. 14-3 [ or Eq. (14-4)]. 
 
Solution: Use mathematical computation: 
 Use Fig. 14-3 or Eq. 14-4 
 q
c
 = (9.39 kW/K)(t
c
 - t
amb
) 
 at 30 C 
 q
c
 = (9.39 kW/K)(t
c
 -30) 
 
 Range of Evaporator Temperature, Fig. 14-1. 
 -10 C, -5 C, 0 C, 5 C, and 10 C. 
 
Eq. (14-1), constant at Table 14-1, Fig. 14-1. 
2
c
2
e9
2
ce8c
2
e7ce6
2
c5c4
2
e3e21e ttcttcttcttctctctctccq ++++++++= 
Eq. (14-2) constant at Table 14-1, Fig. 14-1. 
2
c
2
e9
2
ce8c
2
e7ce6
2
c5c4
2
e3e21 ttdttdttdttdtdtdtdtddP ++++++++= 
Eq. (14-3) 
q
c
 = q
e
 + P 
 
Solving for t
c
 at t
e
 = -10 C 
( ) ( ) 2cc2e 0.001525t1.118157t-100.061652104.60437137.402q −−+−+= 
( ) ( ) ( ) 2cc2c t100.00026682t100.00040148t100.0109119 −−−−−− 
( ) 2c2 t1030.00000387 −+ 
 
2
cce 0.0015305t1.049186t97.5235q +−= 
 
( ) ( ) 2cc2 0.0063397t-0.870024t100.01426100.8932221.00618P +−−−−= 
( ) ( ) ( ) 2cc2c t100.00014746t100.00023875t10-0.033889 −−−−+ 
( ) 2c2 t10620.00000679 −+ 
 
2
cc t0.004185480.507259t8.5124P −+= 
 
q
c
 = q
e
 + P 
 
2
ccc t0.002654980.541927t-106.0359q −= 
Then, 
q
c
 = (9.39 kW/K)(t
c
 -30) 
 
2
ccc t0.002654980.541927t-106.0359281.7-9.39t −= 
0387.73599.931927tt0.00265498 c
2
c =−+ 
 
t
c
 = 38.64 C 
 
( ) ( )2e 38.640.001530538.641.04918697.5235q +−= 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 2 of 11 
q
e
 = 59.3 kW at t
e
 = -10 C. 
 
Solving for t
c
 at t
e
 = -5 C 
( ) ( ) 2cc2e 0.001525t1.118157t-0.0616524.60437137.402q −−+−+= 55 
( ) ( ) ( ) 2cc2c t0.00026682t50.00040148t0.0109119 55 −−−−−− 
( ) 2c2 t30.00000387 5−+ 
 
2
cce 5t0.000094071.0736345t115.92145q +−= 
 
( ) ( ) 2cc2 0.0063397t-0.870024t50.014260.8932221.00618P +−−−−= 5 
( ) ( ) ( ) 2cc2c t0.00014746t0.00023875t5-0.033889 55 −−−−+ 
( ) 2c2 t620.00000679 5−+ 
 
2
cc 5t0.00543249t0.694610255.11579P −+= 
 
q
c
 = q
e
 + P 
 
2
ccc t0.00552657t0.37902425-121.03724q −= 
 
 
 
2
ccc t0.00552657t0.37902425-121.03724281.79.39t −=− 
 
0402.73724t9.76902425t0.00552657 c
2
c =−+ 
 
 t
c
 = 40.31 C 
 
 
( ) ( )2e 40.31750.0000094040.311.0736345115.92145q +−= 
 q
e
 = 72.5 kW at t
e
 = -5 C. 
 
 
 Solvingfor t
c
 at t
e
 = 0 C 
( ) ( ) 2cc2e 0.001525t1.118157t-0.0616524.60437137.402q −++= 00 
( ) ( ) ( ) 2cc2c t0.00026682t0.00040148t0.0109119 000 −−− 
( ) 2c2 t30.00000387 0+ 
 
2
cce 0.0001525t1.118157t137.402q +−= 
 
( ) ( ) 2cc2 0.0063397t-0.870024t00.0142600.8932221.00618P +−−= 
( ) ( ) ( ) 2cc2c t00.00014746t00.00023875t00.033889 −−+ 
( ) 2c2 t0620.00000679+ 
 
2
cc 0.0063397t0.870024t1.00618P −+= 
 
q
c
 = q
e
 + P 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 3 of 11 
 
2
ccc 0.0078647t0.248133t-138.40818q −= 
 
2
ccc 0.0078647t0.248133t-138.40818281.79.39t −=− 
 
0420.108189.638133t0.0078647t c
2
c =−+ 
 
 t
c
 = 42.14 C 
 
 
( ) ( )2e 42.140.00152542.141.118157137.402q +−= 
 q
e
 = 87.6 kW at t
e
 = 0 C. 
 
 
 Solving for t
c
 at t
e
 = 5 C 
( ) ( ) 2cc2e 0.001525t1.118157t-0.0616524.60437137.402q −++= 55 
( ) ( ) ( ) 2cc2c t0.00026682t0.00040148t0.0109119 555 −−− 
( ) 2c2 t30.00000387 5+ 
 
2
cce 5t0.002762271.1827535t161.96515q +−= 
 
( ) ( ) 2cc2 0.0063397t-0.870024t50.0142650.8932221.00618P +−−= 
( ) ( ) ( ) 2cc2c t50.00014746t50.00023875t50.033889 −−+ 
( ) 2c2 t5620.00000679+ 
 
2
cc 5t0.00690709t1.033500253.81643P −+−= 
 
q
c
 = q
e
 + P 
 
2
ccc t0.00966937t0.14925325-158.14872q −= 
 
2
ccc t0.00966937t0.14925325-158.14872281.79.39t −=− 
 
0439.84872t9.53925325t0.00966937 c
2
c =−+ 
 
 t
c
 = 44.14 C 
 
 
( ) ( )2e 44.1450.0027622744.141.1827535161.96515q +−= 
 q
e
 = 104.4 kW at t
e
 = 5 C. 
 
 Solving for t
c
 at t
e
 = 10 C 
( ) ( ) 2cc2e 0.001525t1.118157t-100.061652104.60437137.402q −++= 
( ) ( ) ( ) 2cc2c t100.00026682t100.00040148t100.0109119 −−− 
( ) 2c2 t30.00000387 10+ 
 
2
cce 0.0038059t1.267424t189.6109q +−= 
 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 4 of 11 
( ) ( ) 2cc2 0.0063397t-0.870024t100.01426100.8932221.00618P +−−= 
( ) ( ) ( ) 2cc2c t100.00014746t100.00023875t100.033889 −−+ 
( ) 2c2 t10620.00000679+ 
 
2
cc 5t0.00690709t1.033500253.81643P −+−= 
 
q
c
 = q
e
 + P 
 
2
ccc t0.010940580.082435t-180.25886q −= 
 
2
ccc t0.010940580.082435t-180.25886281.79.39t −=− 
 
0461.958869.472435tt0.01094058 c
2
c =−+ 
 
 t
c
 = 46.29 C 
 
 
( ) ( )2e 46.290.003805946.291.267424189.6109q +−= 
 q
e
 = 122.8 kW at t
e
 = 10 C. 
 
Ans. 
qe, kw 122.8 104.4 87.6 72.5 59.3 
te, C 10 5 0 -5 -10 
tc, C 46.29 44.14 42.14 40.31 38.64 
 
 
14-2. Combine the condensing unit of Problem 14-1 (using answers provided) with the evaporator of Fig. 14-8 to 
form a complete system. The water flow rate to the evaporator is 2 kg/s, and the temperature of water to be 
chilled is 10 C. 
(a) What are the refrigerating capacity and power requirement of this system? 
(b) This system pumps heat between 10 C and an ambient temperature of 30 C, which is the same 
temperature difference as from 15 to 35 C, for which information is available in Table 14-4. Explain 
why the refrigerating capacity and power requirement are less at the lower temperature level. 
 
Solution: 
 
 (a) Eq. 14-6. 
 
( )[ ]( )ewiewie tttt0.04616.0q −−+= 
 t
wi
 = 10 C 
 Expressing q
e
 = f(t
e
) from Problem 14-1. 
 
2
eee 0.03457t3.178t87.5914q ++= 
 Then: 
 
( )[ ]( )eee t10t100.04616.0q −−+= 
 
( )( )eee 0.046t1.466t60q −−= 
 
2
eee 0.276t11.52t87.6q +−= 
 
2
ee
2
ee 0.03457t3.178t87.59140.276t11.52t87.6 ++=+− 
 
00.008614.698t0.24143t e
2
e =+− 
 
C42.14tC,0t ce =≈ 
 Then, q
e
 = 87.6 kw - - - Ans. 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 5 of 11 
 
 
2
cc 0.0063397t0.870024t1.00618P −+= 
 
( ) ( )242.140.006339742.140.8700241.00618P −+=
 
 P = 26.4 kw - - - Ans. 
 q
c
 = q
e
 + p = 87.6 kw + 26.4 kw = 114 kw 
 
(b) At lower temperature level, if t
wi
 = 15 C and ambient temperature= 35 C. 
 From Fig. 14-9. 
 15 C Entering Water Te.mperature 
 35 C Ambient Temperature 
 
 t
e
 = Evaporator Temp = 4.4 C 
 q
e
 = Refrigerating Capacity = 96 kw 
 
 Table 14-3. 
 P = 30 kw 
 t
c
 = 48.4 C 
 q
c
 = 125.8 kw 
 
 Answer. 
 All values above are higher than low temperature level. Therefore refrigerating capacity and power 
are less at low temperature level due to lower ambient temperature and lower entering water temperature to 
be chilled. 
 
14-3. Section 14-11 suggests that the influences of the several components shown in Table 14-6 are dependent 
upon the relative sizes of the components at the base condition. If the base system is the same as that 
tabulated in Table 14-6 except that the condenser is twice as large [ F = 18.78 kW/K in Eq. (14-4)], what is 
the increase in system capacity of a 10 percent increase in condenser capacity above this new base 
condition? The ambient temperature is 35 C, and the entering temperature of the water to be chilled is 15 C. 
 
Solution: 
 35 C ambient temperature, t
amb
. 
 15 C entering temperature of water, t
wi
. 
 
 Eq. 14-4. 
 
( )ambcc t-tFq = 
 
( )35-t18.78q cc = 
 Eq. 14-6 
 
( )[ ]( )ewiewie tttt0.04616.0q −−+= 
 
( )[ ]( )eee t15t150.04616.0q −−+= 
Eq. 14-1. 
2
c
2
e
2
cec
2
ece
2
cc
2
eee
t3t0.00000387tt0.00026682tt0.00040148t0.0109119t
0.001525t1.118157t0.061652t4.60437t137.402q
+−−−
−−++=
 
Eq. 14-2. 
2
c
2
e
2
cec
2
ece
2
cc
2
ee
t62t0.00000679tt0.00014746tt0.00023875t0.033889t
0.0063397t0.870024t0.01426t0.893222t1.00618P
+−−+
−+−−=
 
Eq. 14-3. 
q
c
 = q
e
 + P 
 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 6 of 11 
2
c
2
e
2
cec
2
ece
2
cc
2
eec
t92t0.00001066tt0.00041428tt0.00064023t0.0229771t
0.0078647t0.248133t0.047392t3.711148t138.40818q
+−−+
−−++=
 
 
Use q
c
: 
( )
2
c
2
e
2
cec
2
ece
2
c
c
2
eec
t92t0.00001066
tt0.00041428tt0.00064023t0.0229771t0.0078647t
0.248133t0.047392t3.711148t138.4081835-t18.78
+
−−+−
−++=
 
Let Equation A is, 
0t92t0.00001066tt0.00041428tt0.00064023t0.0229771t
0.0078647t19.028133t0.047392t3.711148t795.70818
2
c
2
e
2
cec
2
ece
2
cc
2
ee
=+−−
+−−++
 
 
Use q
e
: 
 
( )[ ]( ) ( )( )
2
c
2
e
2
cec
2
ece
2
c
c
2
ee
2
ee
eeee
t3t0.00000387
tt0.00026682tt0.00040148t0.0109119t0.001525t
1.118157t0.061652t4.60437t137.4020.276t14.28t152.1
0.046t1.696t90t15t-150.04616.0
+
−−−−
−++=+−
=−==−+
 
 
Let Equation B is, 
14.698t3t0.00000387tt0.00026682tt0.00040148
t0.0109119t0.001525t1.118157t0.214348t18.88437tX
2
c
2
e
2
cec
2
e
ce
2
cc
2
ee
=+−−
−−−−=
 
 Assume a value of t
e
, solve t
c
 from Equation A then substiture in Equaton B to reach 14.698 value. 
 
 Say t
e
 = 0 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t0920.00001066
t00.00041428t00.00064023t00.0229771
0.0078647t19.028133t00.04739203.711148795.70818
2
c
2
2
cc
2
c
2
cc
2
=+
−−
+−−++
 
 
0795.7081819.028133t0.0078647t c
2
c =−+ 
 t
c
 = 41.12 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( ) ( )( )
( ) ( ) 14.69848.5641.12030.00000387
41.1200.0002668241.1200.0004014841.1200.0109119
42.120.00152542.121.11815700.214348018.88437X
22
22
22
〉−=+
−−−
−−−=
 
 
 Say t
e
 = 5 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t5920.00001066
t50.00041428t50.00064023t50.0229771
0.0078647t19.028133t50.04739253.711148795.70818
2
c
2
2
cc
2
c
2
cc
2
=+
−−+−−++
 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 7 of 11 
 
0815.448725t18.9292532t0.00966937 c
2
c =−+ 
 t
c
 = 42.17 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( ) ( )( )
( ) ( ) 14.69842.17530.00000387
42.1750.0002668242.1750.0004014842.1750.0109119
42.170.00152542.171.11815750.214348518.88437X
22
22
22
〉=+
−−−
−−−=
27.34 
 
 
 
 Say t
e
 = 4 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t4920.00001066
t40.00041428t40.00064023t40.0229771
0.0078647t19.028133t40.04739243.711148795.70818
2
c
2
2
cc
2
c
2
cc
2
=+
−−
+−−++
 
 
0811.3110448t18.946468228t0.00935111 c
2
c =−+ 
 t
c
 = 41.95 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( ) ( )( )
( ) ( ) 14.69818.6541.95430.00000387
41.9540.0002668241.9540.0004014841.9540.0109119
41.950.00152541.951.11815740.214348418.88437X
22
22
22
〉=+
−−−
−−−=
 
 
 Say t
e
 = 3.5 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t3.5920.00001066
t3.50.00041428t3.50.00064023t3.50.0229771
0.0078647t19.028133t3.50.0473923.53.711148795.70818
2
c
2
2
cc
2
c
2
cc
2
=+
−−
+−−++
 
 
0809.277757t18.955555923t0.00918398 c
2
c =−+ 
 t
c
 = 41.845 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( ) 14.69810.64541.8453.530.0000038741.8453.50.00026682
41.8453.50.0004014841.8453.50.0109119
41.8450.00152541.8451.1181573.50.2143483.518.88437X
222
2
22
〈=+−
−−
−−−=
 
 
 Say t
e
 = 3.75 C 
Equation A. 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 8 of 11 
( ) ( )
( ) ( ) ( )
( ) 0t3.75920.00001066
t3.750.00041428t3.750.00064023t3.750.0229771
0.0078647t19.028133t3.750.0473923.753.711148795.70818
2
c
2
2
cc
2
c
2
cc
2
=+
−−+
−−++
 
 
0810.2914351t18.95097214375t0.00926821 c
2
c =−+ 
 t
c
 = 41.90 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( )
14.69814.662
41.903.7530.0000038741.903.750.00026682
41.903.750.0004014841.903.750.0109119
41.900.00152541.901.1181573.750.2143483.7518.88437X
222
2
22
≈=
+−
−−
−−−=
 
 
Therefore: t
e
 = 3.75 C and t
c
 = 41.90 C 
 
( ) ( ) ( )
( ) ( )( ) ( ) ( )
( )( ) ( ) ( )222
22
2
e
41.903.7530.0000038741.903.750.00026682
41.903.750.0004014841.903.750.010911941.900.001525
41.901.1181573.750.0616523.754.60437137.402q
+−
−−−
−++=
 
q
e
 = 102.4 kW 
 
or 
( )[ ]( )75.375.3 −−+= 15150.04616.0qe 
q
e
 = 102.4 kW 
 
( ) ( ) ( )
( ) ( )( ) ( ) ( )
( )( ) ( ) ( )222
22
2
41.903.75620.0000067941.903.750.00014746
41.903.750.0002387541.903.750.03388941.900.0063397
41.900.8700243.750.014263.750.8932221.00618P
+−
−+−
+−−=
 
 
P = 27.3 kW 
 
q
c
 = q
e
 + P = 102.4 kW + 27.3 kW = 129.7 kW 
 
 or q
c
 = 18.78 (t
c
 - 35) = 18.78 (41.90 - 35) = 129.6 kW 
 
 New Base Conditions: 
 Compressor = 27.3 kw 
 Condenser = 129.7 kw 
 Evaporator = 102.4 kw 
 
 If condenser capacity is increased by 10 % 
 F = 18.78 x 1.1 = 20.658 
 
 Equation A: 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 9 of 11 
( )
2
c
2
e
2
cec
2
ece
2
c
c
2
eec
t92t0.00001066
tt0.00041428tt0.00064023t0.0229771t0.0078647t
0.248133t0.047392t3.711148t138.4081835-t20.658
+
−−+−
−++=
 
0t92t0.00001066tt0.00041428tt0.00064023t0.0229771t
0.0078647t20.906137t0.047392t3.711148t861.43818
2
c
2
e
2
cec
2
ece
2
cc
2
ee
=+−−
+−−++
 
 
 Say t
e
 = 3.75 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t3.75920.00001066
t3.750.00041428t3.750.00064023t3.750.0229771
0.0078647t20.906137t3.750.0473923.753.711148861.43818
2
c
2
2
cc
2
c
2
cc
2
=+
−−+
−−++
 
 
0876.02143520.828976t44t0.00926821 c
2
c =−+ 
 t
c
 = 41.30 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( )
14.69815.484
41.303.7530.0000038741.303.750.00026682
41.303.750.0004014841.303.750.0109119
41.300.00152541.301.1181573.750.2143483.7518.88437X
222
2
22
≈=
+−
−−
−−−=
 
 
 Say t
e
 = 3.70 C 
Equation A. 
( ) ( )
( ) ( ) ( )
( ) 0t3.7920.00001066
t3.70.00041428t3.70.00064023t3.70.0229771
0.0078647t20.906137t3.70.0473923.73.711148861.43818
2
c
2
2
cc
2
c
2
cc
2
=+
−−+
−−++
 
 
0875.818224t20.8298865465t0.00925147 c
2
c =−+ 
 t
c
 = 41.29 C 
 
Equation B. 
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( )
14.69814.682
41.293.730.0000038741.293.70.00026682
41.293.70.0004014841.293.70.0109119
41.290.00152541.291.1181573.70.2143483.718.88437X
222
2
22
≈=
+−
−−
−−−=
 
 
Then: t
e
 = 3.70 C and t
c
 = 41.29 C 
 
( ) ( ) ( )
( ) ( )( ) ( ) ( )
( )( ) ( ) ( )222
22
2
e
41.293.730.0000038741.293.70.00026682
41.293.70.0004014841.293.70.010911941.290.001525
41.291.1181573.70.0616523.74.60437137.402q
+−
−−−
−++=
 
q
e
 = 103 kW 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 10 of 11 
 
or 
( )[ ]( )3.70153.70150.04616.0qe −−+= 
q
e
 = 103.04 kW 
 
( ) ( ) ( )
( ) ( )( ) ( ) ( )
( )( ) ( ) ( )222
22
2
41.293.7620.0000067941.293.70.00014746
41.293.70.0002387541.293.70.03388941.290.0063397
41.290.8700243.70.014263.70.8932221.00618P
+−
−+−
+−−=
 
 
P = 27.0 kW 
 
 q
c
 = q
e
 + P = 103 kW + 27 kW = 130 kW 
 
 or q
c
 = 20.658 (t
c
 - 35) = 20.658 (41.29 - 35) = 130 kW 
 
 
100%
kw102.4
kw102.4kw103.04
capacitysysteminIncrease ×
−
=
 
 Increase in system capacity = 0.62 % - - - Ans. 
 
 
14-4. For the components of the complete system described in Secs. 14-7, 14-8, and 14-11 the following costs (or 
savings) are applicable to a 1 percent change in component capacity. An optimization is now to proceed by 
increasing or decreasing sizes of components in order to reduce the first cost of the system. What relative 
changes in components sizes should be made in order to reduce the first cost of the system but maintain a 
fixed refrigerating capacity? 
 
 Increase (saving) in first cost 
 Component for 1 % increase (decrease) 
 in component capacity 
 ________________________________________________ 
 Compressor $ 2.80 
 Condenser 0.67 
 Evaporator 1.40 
 
 
Solution: 
 
Tabulation of increase and decrease. 
 
Compressor Condenser Evaporator Total Increase/ 
 Reduction 
-2.80 +0.67 +1.40 -0.73 
-2.80 -0.67 +1.40 -2.07 
-2.80 +0.67 -1.40 -3.53 
-2.80 -0.67 -1.40 -4.87 
+2.80 +0.67 +1.40 +4.87 
+2.80 -0.67 +1.40 +3.53 
+2.80 +0.67 -1.40 +2.07 
+2.80 -0.67 -1.40 +0.73 
 
 The compressor should be increased to avoid freezing of water. So 
try evaporation reduced by 3 % or 2 %. 
 
CHAPTER 14 - VAPOR-COMPRESSION-SYSTEM ANALYSIS 
 Page 11 of 11 
 
Compressor Condenser Evaporator Total Increase/ 
 Reduction 
 
+2.80 +0.67 -3(1.40) -0.73 
+2.80 -0.67 -3(1.40) -2.07 
+2.80 -0.67 -2(1.40) -0.67 > -0.73 
+2.80 +0.00 -3(1.40) -1.40 
 
Answer: Therefore use 3% evaporator capacity decrease for every 1 % increase in compressor capacity 
 
 
- 0 0 0 - 
CHAPTER 15 - REFRIGERANTS 
 Page 1 of 4 
15-1. The machine room housing the compressor and condenser of a refrigerant 12 system has dimensions 5 by 4 
by 3 m. Calculate the mass of the refrigerant which would have to escape into the space to cause a toxic 
concentration for a 2-h exposure. 
 
Solution: 
 Section 15-7, Refrigerant 12 exposure for 2-h has 20 % by volume to become toxic.Room volume = 5 x 4 x 3 m = 60 m
3
. 
 
 Volume of refrigerant 12. 
 = (0.20)(60) = 12 m
2
. 
 
 At atmospheric, 101.325 kPa, Table A-5. 
 ν
g
 = 158.1254 L/kg = 0.1581254 m
3
/kg 
 
 Mass of refrigerant 12. 
 = (12 m
2
) / (0.1581254 m
3
/kg) 
 = 76 kg - - - Ans. 
 
 
15-2. Using data from Table 15-4 for the standard vapor-compression cycle operating with an evaporating 
temperature of -15 C and a condensing temperature of 30 C, calculate the mass flow rate of refrigerant per 
kilowatt of refrigeration and the work of compression for (a) refrigerant 22 and (b) ammonia. 
 
Solution: 
 Table 15-4. 
 (a) Refrigerant 22. 
 
 Suction vapor flow per kW of refrigeration = 0.476 L/s 
 Table A-6, at -15 C evaporating temperature 
 ν
suc
 = 77.68375 L/kg 
 
 mass flow rate = (0.476 L/s) / (77.68375 L/kg) 
 = 0.0061274 kg/s - Ans. 
 
 Work of compression = (mass flow rate)(refrigerating effect) / COP 
 = (0.0061274 kg/s)(162.8 kJ/kg) / 4.66 
 = 0.2141 kW -- - Ans. 
 
 (b) Ammonia (717). 
 
 Suction vapor flow per kW of refrigeration = 0.476 L/s 
 Table A-3, at -15 C evaporating temperature 
 ν
suc
 = 508.013 L/kg 
 
 mass flow rate = (0.476 L/s) / (508.013 L/kg) 
 = 0.00090943 kg/s - Ans. 
 
 Work of compression = (mass flow rate)(refrigerating effect) / COP 
 = (0.00090943 kg/s)(1103.4 kJ/kg) / 4.76 
 = 0.2108 kW -- - Ans. 
 
15-3. A 20% ethylene glycol solution in water is gradually cooled/ 
(a) At what temperature does crystalluzation begin? 
(b) If the antifreeze is cooled to -25 C, what percent will have frozen into ice? 
CHAPTER 15 - REFRIGERANTS 
 Page 2 of 4 
 
Solution: 
 Figure 15-1 and Figure 15-2. 
 
 (a) At point B, 20 % Ethylene Glycol 
 Crystallization Temperature = -8.5 C 
 
 (b) If cooled to -25 C. 
 x
1
 = 0.20 
 x
2
 = 0.425 
 
( )100
xx
x
icePercent
21
1
+
=
 
 
( )100
0.4250.20
0.20
icePercent
+
=
 
 Percent ice = 32 % - - - Ans. 
 
 
15-4. A solution of ethylene glycol and water is to be prepared for a minimum temperature of -30 C. If the 
antifreeze is mixed at 15 C, what is the required specific gravity of the antifreeze solution at this 
temperature? 
 
Solution: 
 Fig. 15-1 and Fig. 15-2 at -30 C, point B 
 concentration = 46 % glycol 
 Figure. 15-3, at 15 C, 46 % glycol. 
 Specific gravity based on water = 1.063 - - - Ans. 
 
15-5. For a refrigeration capacity of 30 kW, how many liters per second of 30 % solution of ethylene glycol-water 
must be circulated if the antifreeze enters the liquid chiller at -5 C and leaves at -10 C? 
 
Solution 
Figure 15-6. 
 At -5 C, cp = specific heat = 3.75 kJ/kg.K 
 At -10 C, cp = specific heat = 3.75 kJ/kg.K 
 
 q = 30 kw = w (3.75 kJ/kg.K)(-5 C - (-10 C)) 
 
 w = 1.60 kg/s 
 
 Specific gravity at -7.5 C = 1.0475 
 
 Liters per second = (1.60 kg/s)(1 / 1.0475 kg/L) 
 Liters per second = 1.53 L/s - - - Ans. 
 
 
15-6. A manufacturer’s catalog gives the pressure drop through the tubes of a heat-exchanger as 70 kPa for a 
given flow rate of water at 15 C. If a 40 % ethylene glycol-water solution at -20 C flows through the heat 
exchanger at the same mass flow rate as the water, what will the pressure drop be? Assume turbulent flow. 
At 15 C the viscosity of water is 0.00116 Pa/.s. 
 
Solution: 
 Equation 15-3. 
CHAPTER 15 - REFRIGERANTS 
 Page 3 of 4 
 
w
w
w
w
w
a
a
a
a
a
w
a
2
2V
D
L
f
2
2V
D
L
f
p
p
ρ
ρ
=
∆
∆
 
 Equation 15-4. 
 
0.25Re
0.33
f =
 
 
 
µ
ρ
=
DV
Re
 
 ∆p
w
 = 70 kPa 
 µ
w
 = 0.0016 Pa.s 
 ρ
w
 = 0.99915 kg/L at 15 C 
 
 
wa
w
w
a
a DD,
D
L
D
L
==
 
 
0.25
aaw
wwa
0.25
a
0.25
w
w
a
V
V
Re
Re
f
f








ρµ
ρµ
==
 
 But: 
 a
a
w
w
A
w
V;
A
w
V
ρ
=
ρ
=
 
 Then: 
 
0.25
w
a
w
a
f
f






µ
µ
=
 
 Equation 15-3 then becomes, 
 
2
w
a
w
a
0.25
w
a
w
a
V
V
∆p
∆p












ρ
ρ






µ
µ
=
 
 
2
a
w
w
a
0.25
w
a
w
a
∆p
∆p








ρ
ρ






ρ
ρ






µ
µ
=
 
 








ρ
ρ






µ
µ
=
a
w
0.25
w
a
w
a
∆p
∆p
 
 For 40 % Ethylene Glycol, -20 C. 
 Fig. 15-3, Specific Gravity = 1.069 
 
 ρ
a
 = 1.069 kg/L 
 
 Fig. 15-5 
 
 µ
a
 = 0.01884 Pa.s 
 
 Substitute: 
 
 












=
1.069
0.99915
0.00116
0.01884
70
∆p
0.25
a
 
CHAPTER 15 - REFRIGERANTS 
 Page 4 of 4 
 ∆∆∆∆pa = 131 kPa - - - Ans. 
 
15-7. Compute the convection heat-transfer coefficient for liquid flowing through a 20-mm-ID tube when the 
velocity is 2.5 m/s if the liquid is (a) water at 15 C, which has a viscosity of 0.00116 Pa.s and a thermal 
conductivity of 0.584 W/m.K; (b) 40 % solution of ethylene glycol at -20 C. 
 
Solution: 
 Equation 15-5. 
 
0.4
p
0.8
k
cVD
D
k
0.023h 






 µ






µ
ρ
=
 
 (a) Water: 
 
 ρ = 0.99915 kg/L = 999.15 kg/m3 
 D = 0.020 m 
 µ = 0.00116 Pa.s 
 k = 0.584 W/m.K 
 c
p
 = 4190 J/kg.K 
 V = 2.5 m/s 
 
( )( )( ) ( )( ) 0.40.8
0.584
0.001164190
0.0016
999.150.0202.5
0.020
0.584
0.023h 

















=
 
 h = 6,177 W/m
2
.K - - - Ans. 
 
 (b) 40 % Solution, Ethylene Glycol at -20 C 
 
 ρ = 1.069 kg/L (Fig. 15-3) = 1069 kg/m3 
 D = 0.020 m 
 µ = 0.01884 Pa.s (Fig. 15-5) 
 k = 0.45 W/m.K (Fig. 15-4) 
 c
p
 = 3450 J/kg.K (Fig. 15-6) 
 V = 2.5 m/s 
 
( )( )( ) ( )( ) 0.40.8
0.450
0.018843450
0.01884
10690.0202.5
0.020
0.450
0.023h 

















=
 
 h = 2,188 W/m
2
.K - - - Ans. 
 
- 0 0 0 - 
 
 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 1 of 9 
16-1 A cylindrical tank 2 m long mounted with its axis horizontal is to separate liquid ammonia from ammonia 
vapor. The ammonia vapor bubbles through the liquid and 1.2 m
3
/s leaves the surface of the liquid. If the 
velocity of the vapor is limited to 1 m/s and the vessel is to operate with the liquid level two-thirds of the 
diameter from the bottom, what must the diameter of the tank be? 
 
Solution: 
 
 L = 2 m 
 Surface Area = A = (1.2 m
3
/s) / (1 m/s) = 1.2 m
2
 
 Width = W = A/L = (1.2 m
2
) / (2 m) = 0.6 m 
 
 
 
xD
2
1
D
3
2
+=
 
 
22 WD
2
1
x −=
 
 
22 WD
2
1
D
2
1
D
3
2
−+=
 
 
22 WDD
3
1
−=
 
 
( )222 0.6WD
9
8
==
 
 D = 0.636 m - - - Ans. 
 
16-2. A liquid subcooler as shown in Fig. 16-14 receives liquid ammonia at 30 C and subcools 0.6 kg/s to 5 C. 
Saturated vapor leaves the subcooler for the high-stage compressor at -1 C. Calculate the flow rate of 
ammonia that evaporated to cool the liquid. 
 
Solution: Refer to Fig. 16-14. 
 
 Liquid ammonia at 30 C, Table A-3. 
 h
1
 = h
f
 = 341.769 kJ.kg 
 Subcooled ammonia at 5 C, Table A-3. 
 h
2
 = h
f
 = 223.185 kJ/kg 
 Saturated vapor ammonia at -1 C, Table A-3. 
 h
3
 = h
g
 = 1460.62 kJ/kg 
 
 Heat Balance: 
 
 w
1
(h
1
 - h
2
) = w
2
 (h
3
 - h
1
) 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 2 of 9 
 (0.6)(341.769 - 223.185) = w
2
 (1460.62- 341.769) 
 
 w
2
 = 0.0636 kg/s - - - Ans. 
 
 
16-3. In a refrigerant 22 refrigeration system the capacity is 180 kw at a temperature of -30 C. The vapor from the 
evaporator is pumped by one compressor to the condensing pressure of 1500 kPa. Later the system is 
revised to a two-stage compression operating on the cycle shown in Fig. 16-6 with intercooling but no 
removal of flash gas at 600 kPa. 
(a) Calculate the power required by the single compressor in the original system. 
(b) Calculate the power required by the two compressor in the revised system. 
 
Solution: 
 
 (a) Original system 
 
 
 At 1, -30 C, Table A-6. 
 h
1
 = 393.138 kJ/kg 
 s
1
 = 1.80329 kJ/kg.K 
 
 At 2, 1500 kPa condensing pressure = 39.095 C condensing temp. 
 Table A-7, constant entropy 
 h
2
 = 450.379 kJ/kg 
 
 h
3
 = h
4
 = 248.486 kJ/kg 
 
 w = 180 kw / (h
1
 - h
4
) 
 w = 180 / (393.138 - 248.486) 
 w = 1.2444 kg/s 
 
 P = w (h
2
 - h
1
) 
 P = 1.2444 (450.379 - 393.138) 
 P = 71.23 kw - - - Ans. 
 
 (b) Revised system (Fig. 16-6). 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 3 of 9 
 
 
 At 1, -30 C, Table A-6 
 h
1
 = 393.138 kJ/kg 
 s
1
 = 1.80329 kJ/kg.K 
 
 At 2, 600 kPa, Sat. Temp. = 5.877 C (Table A-7) 
 Constant Entropy 
 h
2
 = 424.848 kJ/kg 
 
 At 3, 600 kPa, Sat. Temp. = 5.877 C (Table A-6) 
 h
3
 = 407.446 kJ/kg 
 s
3
 = 1.74341 kJ/kg.K 
 
 At 4, 1500 kPa, Sat. Temp. = 39.095 C (Table A-7) 
 h
4
 = 430.094 kJ/kg 
 
 At 5, 1500 kPa, Sat. Temp. = 39.095 C (Table A-6) 
 h
7
 = h
6
 = h
5
 = 248.486 kJ/kg 
 
 
 w
1
 = entering low-stage compressor 
 
 w
1
= 180 kw / (h
1
 - h
7
) = 180 / (393.138 - 248.486) 
 w
1
 = 1.2444 kg/s 
 
 w
2
 = enteirng intercooler 
 w
3
 = entering high-stage compressor 
 
 Heat Balance through intercooler 
 
 w
2
h
6
 + w
1
h
2
 = w
3
h
3
 
 
 Mass Balance through intercooler 
 w
2
 + w
1
 = w
3
 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 4 of 9 
 w
2
 + 1.2444 = w
3
 
 w
2
 = w
3
 - 1.2444 
 
 (w
3
 - 1.2444)(248.486) + (1.2444)(424.848) = w
3
 (407.446) 
 w
3
 = 1.38063 kg/s 
 
 P = w
1
(h
2
 - h
1
) + w
3
(h
4
 - h
3
) 
 P = (1.2444)(424.848 - 393.138) + (1.38063)(430.094 - 407.446) 
 P = 70.73 kw - - - Ans. 
 
16-4. A refrigerant 22 system has a capacity of 180 kw of an evaporating temperature of -30 C when the 
condensing pressure is 1500 kPa. 
(a) Compute the power requirement for a system with a single compressor. 
(b) Compute the total power required by the two compressors in the system shown in Fig. 16-7 where 
there is no intercooling but there is flash-gas removal at 600 kPa? 
 
Solution: 
 
 
 (a) Original system 
 
 
 
 At 1, -30 C, Table A-6. 
 h
1
 = 393.138 kJ/kg 
 s
1
 = 1.80329 kJ/kg.K 
 
 At 2, 1500 kPa condensing pressure = 39.095 C condensing temp. 
 Table A-7, constant entropy 
 h
2
 = 450.379 kJ/kg 
 
 h
3
 = h
4
 = 248.486 kJ/kg 
 
 w = 180 kw / (h
1
 - h
4
) 
 w = 180 / (393.138 - 248.486) 
 w = 1.2444 kg/s 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 5 of 9 
 
 P = w (h
2
 - h
1
) 
 P = 1.2444 (450.379 - 393.138) 
 P = 71.23 kw - - - Ans. 
 
 
 (b) For Fig. 16-7. 
 
 
 
 At 1, -30 C, Table A-6 
 h
1
 = 393.138 kJ/kg 
 s
1
 = 1.80329 kJ/kg.K 
 
 At 2, 1500 kPa, Sat. Temp. = 39.095 C (Table A-7) 
 Constant Entropy 
 h
2
 = 450.379 kJ/kg 
 
 At 3, 600 kPa, Sat. Temp. = 5.877 C (Table A-6) 
 h
3
 = 407.446 kJ/kg 
 s
3
 = 1.74341 kJ/kg.K 
 
 At 4, 1500 kPa, Sat. Temp. = 39.095 C (Table A-7) 
 h
4
 = 430.094 kJ/kg 
 
 At 5, 1500 kPa, Sat. Temp. = 39.095 C (Table A-6) 
 h
5
 = 248.486 kJ/kg 
 
 At 7, 600 kPa, Say. Temp. = 5.877 C (Table A-6) 
 h
7
 = 206.943 kJ/kg 
 
 h
6
 = h
5
 = 248.486 kJ/kg 
 h
8
 = h
7
 = 206.943 kJ/kg 
 
 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 6 of 9 
 w
1
 = entering evaporator compressor 
 
 w
1
= 180 kw / (h
1
 - h
8
) = 180 / (393.138 - 206.943) 
 w
1
 = 0.96673 kg/s 
 
 w
2
 = entering flashtank 
 w
3
 = entering flash-gas compressor 
 
 Heat Balance through intercooler 
 
 w
2
h
6
 = w
1
h
7
 + w
3
h
3
 
 
 Mass Balance through intercooler 
 w
2
 = w
1
 + w
3
 
 w
2
 = 0.96673 + w
3
 
 w
3
 = 0.96673 + w
3
 
 
 (w
3
 + 0.96673)(248.486) = (0.96673)(206.943) + w
3
(407.446) 
 w
3
 = 0.25265 kg/s 
 
 P = w
1
(h
2
 - h
1
) + w
3
(h
4
 - h
3
) 
 P = (0.96673)(450.3798 - 393.138) + (0.25265)(430.094 - 407.446) 
 P = 61.06 kw - - - Ans. 
 
16-5. A two-stage ammonia system using flash-gas removal and intercooling operates on the cycle shown in Fig. 
16-12a. The condensing temperature is 35 C. The saturation temperature of the intermediate-temperature 
evaporator is 0 C, and its capacity is 150 kW. The saturation temperature of the low-temperature evaporator 
is -40 C, and its capacity is 250 kW. What is the rate of refrigerant compressed by the high-stage 
compressor? 
 
Solution: 
 
 Refer to Fig. 16-12a. 
 
 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 7 of 9 
 
 At 1, -40 C, Table A-3. 
 h
1
 = 1407.26 kJ/kg 
 s
1
 = 6.2410 kJ/kg.K 
 
 At 2, 0 C, Fig. A-1, Constant Entropy 
 h
2
 = 1666 kJ/kg 
 
 At 3, 0 C, Table A-3 
 h
3
 = 1461.70 kJ/kg 
 
 At 4, 35 C, Fig. A-1 
 h
4
 = 1622 kJ/kg 
 
 At 5, 35 C, Table A-3. 
 h
5
 = 366.072 kJ/kg 
 
 At 6, h
6
 = h
5
 = 366.072 kJ/kg 
 
 At 7, 0 C, Table A-3 
 h
7
 = 200 kJ/kg 
 
 At 8, h
8
 = h
7
 = 200 kJ/kg. 
 
 w
1
 = entering low-stage compressor 
 w
1
 = 250 / (h
1
 - h
8
) 
 w
1
 = 250 / (1407.26 - 200) 
 w
1
 = 0.2071 kg/s 
 
 w
2
 = entering high-stage compressor leaving intercooler and flashtank 
 
 Heat balance through intercooler and flashtank. 
 
 w
2
(h
3
 - h
6
) = w
1
(h
2
 - h
7
) 
 w
2
(1461.70 - 366.072) = (0.2071)(1666 - 200) 
 w
2
 = 0.2771 kg/s 
 
 w
3
 = entering intermediate temperature evaporator 
 
 w
3
 = 150 kw / (h
3
 - h
6
) = 150 / (1461.70 - 366.072) 
 w
3
 = 0.1369 kg/s 
 
 Total refrigerant compressed by high=pressure compressor 
 = w
2
 + w
3
 = 0.2771 + 0.1369 
 = 0.4140 kg/s - - - Ans. 
 
16-6. A two-stage refrigerant 22 system that uses flash-gas removal and intercooling serves a single low-
temperature evaporator, as in Fig. 16-10a. The evaporator temperature is -40 C, and the condensing 
temperature is 30 C. The pumping capacity of the high- and low-stage compressors is shown in Fig. 16-18. 
What is (a) the refrigerating capacity of the system and (b) the intermediate pressure? 
CHAPTER 16 - MULTI PRESSURE SYSTEMS 
 Page 8 of 9 
 
Solution: Refer to Fig. 16-18 and Fig. 16-10a. 
 
 
 
 At 1, -40 C, Table A-6 
 h
1
 = 388.609 kJ/kg 
 s
1
 = 1.82504 kJ/kg.K 
 At 5, 30 C, Table A-6 
 h
5
 = 236.664 kJ/kg 
 
 Trial 1 
 
( )( ) kPa3541192105p i == 
 At 354 kPa, Sat. Temp. = -10 C 
 
 At 2, -10 C, Constant Entropy, Table A-7 
 h
2
 = 417.46 kJ/kg 
 
 At 3, -10 C, Table A-6 
 h
3
 = 401.555 kJ/kg 
 s
3
 = 1.76713 kJ/kg.K 
 
 At 4, 30 C, Constant Entropy, table A-7 
 h
4
 = 431.787 kJ/kg 
 
 At 6, h6 = h5 = 236.664 kJ/kg 
 At 7, -10 C, Table A-6 
 h
7
 = 188.426 kJ/kg 
 
 At 8, h
8
 = h
7
 - 188.426 kJ/kg 
 
 w
1
 = low-stage compressor 
 w
2
 = high-stage compressor 
 
CHAPTER 16 - MULTI PRESSURESYSTEMS 
 Page 9 of 9 
 w
1
(h
2
 - h
7
) = w
2
(h
3
 - h
6
) 
 
 
1.3891.139
236.664401.555
188.426417.46
hh
hh
w
w
63
72
1
2 <=
−
−
=
−
−
=
 
 Figure 16-18, at 354 kPa. 
 w
1
 = 1.8 kg/s 
 w
2
 = 2.05 kg/s 
 w
2
/w
1
 = 2.05 / 1.8 = 1.139 < 1.389 
 
 Next trial: p
i
 = 390 kPa 
 
 At 390 kPa, Sat. Temp. = -7.26 C. 
 
 At 2, -7.26 C, Constant entropy, Table A-7 
 h
2
 = 419.836 kJ/kg 
 
 At 3, -7.26 C, Table A-6 
 h
3
 = 402.629 kJ/kg 
 s
3
 = 1.762776 kJ/kg.K 
 
 At 4, 30 C, Constant entropy, Table A-7 
 h
4
 = 430.386 kJ/kg 
 
 At 6, h6 = h5 = 236.664 kJ/kg 
 
 At 7, -7.26 C, Table A-6. 
 h
7
 = 191.570 kJ/kg 
 
 At 8, h
8
 = h
7
 = 191.570 kJ/kg 
 
 
1.3754
236.664402.629
191.570419.836
hh
hh
w
w
63
72
1
2
=
−
−
=
−
−
=
 
 
 Figure 16-18, At 390 kPa. 
 w
1
 = 1.615 kg/s 
 w
2
 = 2.2 kg/s 
 
 
1.37541.36
1.615
2.2
w
w
1
2
≈==
 
 
 Therefore use p
i
 = 390 kPa 
 
 (a) q
e
 = w1(h1 - h8) 
 q
e
 = (1.615)(388.609 - 191.570) 
 q
e
 = 318 kw - - - Ans. 
 
 (b) p
i
 = 390 kPa - - - Ans. 
 
- 0 0 0 - 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 1 of 11 
17-1. What is the COP of an ideal heat-operated refrigeration cycle that receives the energizing heat from a solar 
collector at a temperature of 70 C, performs refrigeration at 15 C, and rejects heat to atmosphere at a 
temperature of 35 C? 
 
Solution: 
 Eq. 17-4. 
 
( )
( )ras
asr
TTT
TTT
COP
−
−
=
 
 Ts = 70 C + 273 = 343 K 
 Tr = 15 C + 273 = 288 K 
 Ta = 35 C + 273 = 308 K 
 
 
( )( )
( )( )288308343
308343288
COP
−
−
=
 
 COP = 1.47 - - -Ans. 
 
17-2. The LiBr-Water absorption cycle shown in Fig. 17-2 operates at the following temperatures: generator, 105 
C; condenser, 35 C; evaporator, 5 C; and absorber, 30 C. The flow rate of solution delivered by the pump is 
0.4 kg/s. 
(a) What are the mass flow rates of solution returning from the generator to the absorber and of the 
refrigerant? 
(b) What are the rates of heat transfer of each component, and the COP
abs
? 
 
Solution: 
 Saturation pressure at 35 C water = 5.63 kPa (condenser) 
 Saturation pressure at 5 C water = 0.874 kPa (evaporator) 
 
 (a) At the generator, LiBr-Water Solution: 
 Fig. 17-5, 105 C, 5.63 kPa, Refer to Fig. 17-2. 
 x
2
 = 70 % 
 
 At the absorber, LiBr-Water 
 Fig. 17-5, 30 C, 0.874 kPa 
 x
1
 = 54 % 
 
 w
1
 = LiBr-Water Solution delivered by pump. 
 w
2
 = Solution returning from generator to absorber. 
 w
3
 = refrigerant water flow rate. 
 
 Total mass-flow balance: 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
 
 LiBr Balance: 
 w
1
x
1
 = w
2
x
2
 
 (0.40)(0.54)= (w
2
)(0.70) 
 w
2
 = 0.3086 kg/s 
 
 Flow rate of solution = w
2
 = 0.3086 kg/s - - - Ans. 
 Flow rate of refrigerant = w
3
 = w
1
 - w
2
 
 w
3
 = 0.40 - 0.3086 
 w
3
 = 0.0914 kg/s - - - Ans. 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 2 of 11 
 
 (b) Refer to Fig. 17-6. 
 
 Enthalpies: 
 Enthalpies of solution, Fig. 17-8. 
 h
1
 = h at 30 C and x of 54 % = -178 kJ/kg 
 h
2
 = h at 105 C and x of 70 % = -46 kJ/kg 
 
 Enthalpy of water liquid and vapor: Table A-2 
 h
3
 = h of saturated vapor at 105 C = 2683.75 kJ/kg 
 h
4
 = h of saturated liquid at 35 C = 146.56 kJ/kg 
 h
5
 = h of saturated liquid at 5 C = 2510.75 kJ/kg 
 
 w
3
=w
4
=w
5
=w
c
 
 
 Generator 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.0914)(2683.75) + (0.3086)(-46) - (0.40)(-178) 
 q
g
 = 302.3 kW - - Ans. 
 
 Condenser 
 q
c
 = w
c
h
3
 - w
4
h
4
 
 q
c
 = (0.0914)(2683.75 - 146.56) 
 q
c
 = 231.9 kW - - Ans. 
 
 Absorber 
 q
a
 = w
2
h
2
+ w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.3086)(-46) + (0.0914)(2510.75) - (0.4)(-178) 
 q
a
 = 286.5 kw - - - Ans. 
 
 Evaporator 
 q
e
 = w5h5 - w4h4 
 q
e
 = (0.0914)(2510.75 - 146.56) 
 q
e
 = 216.1 kW - - - Ans. 
 
 COP = qe / qg = (216.1 kW) / (302.3 kW) 
 COP = 0.715 - - - Ans. 
 
17-3. In the absorption cycle shown in Fig. 17-9 the solution temperature leaving the heat exchanger and entering 
the generator is 48 C. All other temperatures and the flow rate are as shown in Fig. 17-9. What are the rates 
of heat transfer at the generator and the temperature at point 4? 
 
Solution: Refer to Fig. 17-9. 
 w
1
 = w
2
 = 0.6 kg/s 
 w
3
 = w
4
 = 0.452 kg/s 
 
 Heat balance through heat exchanger 
 w
3
h
3
 - w
4
h
4
 = w
2
h
2
 - w
1
h
1
 
 w
3
(h
3
 - h
4
) = w
1
(h
2
-h
1
) 
 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 3 of 11 
 Enthalpies remain unchanged from Ex. 17-4 and Ex. 17-3. 
 h
1
 = -168 kJ/kg 
 h
3
 = -52 kJ/kg 
 
 At point 2, temperature = 48 C 
 Fig. 17-8, x
1
 = 50 % solution, 48 C 
 h
2
 = -128 kJ/kg 
 
 w
3
(h
3
 - h
4
) = w
1
(h
2
-h
1
) 
 (0.452)(-52-h
4
) = (0.6)(-128-(-168)) 
 h
4
 = -105.1 kJ/kg 
 
 q
g
 = w
3
h
3
 + w5h5 - w2h2 
 
 w
5
 = 0.148 kg/s 
 h
5
 = 2676.0 kJ/kg 
 
 q
g
 = (0.452)(-52) + (0.148)(2676) - (0.6)(-128) 
 q
g
 = 449.4 kW - - - Ans. 
 
 At point 4, h
4
 = -105.1 kJ/kg, x
3
 = 66.4 % 
 Fig. 17-8. 
 t
4
 = 70 C - - - Ans. 
 
17-4. The solution leaving the heat exchanger and returning to the absorber is at a temperature of 60 C. The 
generator temperature is 95 C. What is the minimum condensing temperature permitted in order to prevent 
crystallization in the system? 
 
Solution: Refer to Fig. 1709. 
 
 Figure 17-8. 
 At crystaliization, 60 C solution temperature 
 Percent lithium bromide = 66.4 % 
 
 Figure 17-5, x = 66.4 %, 95 C 
 
 Vapor pressure = 6.28 kPa 
 Sat. Temp. of pure water = 37 C 
 
 Minimum condensing temperature = 37 C - - - Ans. 
 
17-5. One of the methods of capacity control described in Sec. 17-11 is to reduce the flow rate of solution delivered 
by the pump: The first-order approximation is that the refrigerating capacity will be reduced by the same 
percentage as the solution flow rate. There are secondary effects also, because if the mean temperature of 
the heating medium in the generator, the cooling water in the absorber and condenser and the water being 
chilled in the evaporator all remain constant, the temperatures in these components will change when the 
heat-transfer rate decreases. 
(a) Fill out each block in the Table 17-1 with either “increases” or “decreases” to indicate qualitative 
influence of the secondary effect. 
(b) Use the expression for an ideal heat-operated cycle to evaluate the effects of temperature on the 
COP
abs
. 
 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 4 of 11 
Solution: 
 
 Use Data of Ex. 17-3 and Ex. 17-2 and Fig. 17-6. 
 
 (a) Initial: 
 w
1
 = 0.6 kg/s 
 w
2
 = 0.452 kg/s 
 w
3
 = w
4
 = w
5
 = 0.148 kg/s 
 x
1
 = 50 % 
 x
2
 = 66.4 % 
 Enthalpies: Fig. 17-8. 
 h
1
 = h at 30 C and x of 50 % = -168 kJ/kg 
 h
2
 = h at 100 C and x of 60 % = -52 kJ/kg 
 Enthalpies: Table A-1 
 h
3
 = h of saturated vapor at 100 C = 2676.0 kJ/kg 
 h
4
 = h of saturated liquid at 40 C = 167.5 kJ/kg 
 h
5
 = h of saturated vapor at 10 C = 2520.0 kJ/kg 
 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 = 473.3 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 = 371.2 kW 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 =450.3 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 = 348.2 kW 
 
 
0.736
q
q
COP
g
e
abs ==
 
 New Solution: 
 When w
1
 is reduced to 0.4 kg/s (concentration of solution remains unchanged as first 
approximation) 
 w
1
 = 0.4 kg/s 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
 w
1
x
1
 = w
2
x
2
 
 (0.4)(0.5) = w
2
(0.664) 
 w
2
 = 0.3012 kg/s 
 w
3
 = 0.0988 kg/s 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.0988)(2676.0) + (0.3012)(-52) - (0.4)(-168) = 315.9 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 q
c
 = (0.0988)(2676.0 - 167.5) = 247.8 kW 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.3012)(-52) + (0.0988)(2520) - (0.4)(-168) = 300.5 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 
 q
e
 = (0.0988)(2520.0 - 167.5) = 232.4 kW 
 
] Assume: Mean temperature of heating medium in the generator = 120 C. Mean temperature of the cooling 
water in the absorber and condenser = 25 C. Mean temperature of the water being chilled in the evaporator = 
15 C. 
 
 New temperature of components: 
 Generator = 120 - (315.9 / 473.3)(120 - 100) = 106.6 C (increase) 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 5 of 11 
 Absorber = 25 + (300.5 / 450.3)(30 - 25) = 28.34 C (decrease) 
 Condenser = 25 + (247.8 / 371.2)(40 - 25) = 35.0 C (decrease) 
 Evaporator = 15 - (233.4 / 348.2)(15 - 10) = 11.66 C (increase) 
 
 With change in component temperature. 
 Fig. 17-5, 35 C condenser temperature, 106.6 C solution temperature 
 x
2
 = 0.70 (increase) 
 At 11.66 C evaporator temperature, 28.34 C solution temperature 
 x
1
 = 0.46 (decrease) 
 
 Enthalpies: Fig. 17-8. 
 h
1
 = h at 28.34 C and x of 46 % = -158 kJ/kg 
 h
2
 = h at 106.6 C and x of 70 % = -45 kJ/kg 
 
 Enthalpies: Table A-1. 
 h
3
 = h of saturated vapor at 106.6 C = 2686.2 kJ/kg 
 h
4
 = h of saturated liquid at 35 C = 146.56 kJ/kg 
 h
5
 = h of saturated vapor at 11.66 C = 2523.0 kJ/kg 
 
 w
1
 = 0.4 kg/s 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
 w
1
x
1
 = w
2
x
2
 
 (0.4)(0.46) = w
2
(0.70) 
 w
2
 = 0.263 kg/s 
 w
3
 = 0.137 kg/s 
 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.137)(2686.2) + (0.263)(-45) - (0.4)(-158) = 419.4 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 
 q
c
 = (0.137)(2686.2 - 146.56) = 348 kW 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.263)(-45) + (0.137)(2523) - (0.4)(-158) = 397 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 
 q
e
 = (0.137)(2523.0 - 146.56) = 325.6 kW 
 
(increase) 0.776
q
q
COP
g
e
abs ==
 
 New temperature of components: 
 Generator = 120 - (419.4 / 473.3)(120 - 100) = 102.3 C (increase) 
 Absorber = 25 + (397 / 450.3)(30 - 25) = 29.4 C (decrease) 
 Condenser = 25 + (348 / 371.2)(40 - 25) = 39.1 C (decrease) 
 Evaporator = 15 - (325.6 / 348.2)(15 - 10) = 10.3 C (increase) 
 
 With change in component temperature. 
 Fig. 17-5, 35 C condenser temperature, 102.3 C solution temperature 
 x
2
 = 0.675 (increase) 
 At 10.3 C evaporator temperature, 29.4 C solution temperature 
 x
1
 = 0.4875 (decrease) 
 
 Enthalpies: Fig. 17-8. 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 6 of 11 
 h
1
 = h at 29.4 C and x of 48.75 % = -165 kJ/kg 
 h
2
 = h at 102.3 C and x of 67.5 % = -50 kJ/kg 
 
 Enthalpies: Table A-1. 
 h
3
 = h of saturated vapor at 102.3 C = 2679.6 kJ/kg 
 h
4
 = h of saturated liquid at 39.1 C = 163.7 kJ/kg 
 h
5
 = h of saturated vapor at 10.3 C = 2520.5 kJ/kg 
 
 w
1
 = 0.4 kg/s 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
 w
1
x
1
 = w
2
x
2
 
 (0.4)(0.4875) = w
2
(0.675) 
 w
2
 = 0.2889 kg/s 
 w
3
 = 0.1111 kg/s 
 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.1111)(2679.6) + (0.2889)(-50) - (0.4)(-165) = 349.3 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 
 q
c
 = (0.1111)(2679.6 - 163.7) = 279.5 kW 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.2889)(-50) + (0.1111)(2520.5) - (0.4)(-165) = 331.6 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 
 q
e
 = (0.1111)(2520.5 - 163.7) = 261.8 kW 
 
(increase) 0.749
q
q
COP
g
e
abs ==
 
 New temperature of components: 
 Generator = 120 - (349.3 / 473.3)(120 - 100) = 105.2 C (increase) 
 Absorber = 25 + (331.6 / 450.3)(30 - 25) = 28.7 C (decrease) 
 Condenser = 25 + (279.5 / 371.2)(40 - 25) = 36.3 C (decrease) 
 Evaporator = 15 - (261.8 / 348.2)(15 - 10) = 11.24 C (increase) 
 
 With change in component temperature. 
 Fig. 17-5, 36.3 C condenser temperature, 105.2 C solution temperature 
 x
2
 = 0.6975 (increase) 
 At 11.24 C evaporator temperature, 28.7 C solution temperature 
 x
1
 = 0.475 (decrease) 
 
 Enthalpies: Fig. 17-8. 
 h
1
 = h at 28.7 C and x of 47.5 % = -162 kJ/kg 
 h
2
 = h at 105.2 C and x of 69.75 % = -45 kJ/kg 
 
 Enthalpies: Table A-1. 
 h
3
 = h of saturated vapor at 105.2 C = 2684.1 kJ/kg 
 h
4
 = h of saturated liquid at 36.3 C = 152 kJ/kg 
 h
5
 = h of saturated vapor at 11.24 C = 2522.2 kJ/kg 
 
 w
1
 = 0.4 kg/s 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 7 of 11 
 w
1
x
1
 = w
2
x
2
 
 (0.4)(0.475) = w
2
(0.6975) 
 w
2
 = 0.2724 kg/s 
 w
3
 = 0.1276 kg/s 
 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.1276)(2684.1) + (0.2724)(-45) - (0.4)(-162) = 395 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 
 q
c
 = (0.1276)(2684.1 - 152) = 323 kW 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.2724)(-45) + (0.1276)(2522.2) - (0.4)(-162) = 374.4 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 
 q
e
 = (0.1276)(2522.2 - 152) = 302.4 kW 
 
(increase) 0.766
q
q
COP
g
e
abs ==
 
 New temperature of components: 
 Generator = 120 - (395 / 473.3)(120 - 100) = 103.3 C (increase) 
 Absorber = 25 + (374.4 / 450.3)(30 - 25) = 29.2 C (decrease) 
 Condenser = 25 + (323 / 371.2)(40 - 25) = 38.1 C (decrease) 
 Evaporator = 15 - (302.4 / 348.2)(15 - 10) = 10.66 C (increase) 
 
 With change in component temperature. 
 Fig. 17-5, 38.1 C condenser temperature, 103.3 C solution temperature 
 x
2
 = 0.675 (increase) 
 At 10.66 C evaporator temperature, 29.2 C solution temperature 
 x
1
 = 0.4875 (decrease) 
 
 Enthalpies: Fig. 17-8. 
 h
1
 = h at 29.2 C and x of 48.75 % = -165 kJ/kg 
 h
2
 = h at 103.3 C and x of 67.5 % = -50 kJ/kg 
 
 Enthalpies: Table A-1. 
 h
3
 = h of saturated vapor at 103.3 C = 2681 kJ/kg 
 h
4
 = h of saturated liquid at 38.1 C = 159.5 kJ/kg 
 h
5
 = h of saturated vapor at 10.66 C = 2521 kJ/kg 
 
 w
1
 = 0.4 kg/s 
 w
2
 + w
3
 = w
1
 = 0.4 kg/s 
 w
1
x
1
 = w
2
x
2
 
 (0.4)(0.4875) = w
2
(0.675) 
 w
2
 = 0.2889 kg/s 
 w
3
 = 0.1111 kg/s 
 
 q
g
 = w
3
h
3
 + w
2
h
2
 - w
1
h
1
 
 q
g
 = (0.1111)(2681) + (0.2889)(-50) - (0.4)(-165) = 349.4 kW 
 q
c
 = w
c
h
3
 - w
4
h
4
 
 q
c
 = (0.1111)(2681 - 159.5) = 280 kW 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 8 of 11 
 q
a
 = w
2
h
2
 + w
5
h
5
 - w
1
h
1
 
 q
a
 = (0.2889)(-50) + (0.1111)(2521) - (0.4)(-165) = 331.6 kW 
 q
e
 = w
5
h
5
 - w
4
h
4
 
 q
e
 = (0.1111)(2521 - 159.5) = 262.4 kW 
 
(increase) 0.751
q
q
COP
g
e
abs ==
 
 New temperature of components: 
 Generator = 120 - (349.4 / 473.3)(120 - 100) = 105.2 C (increase) 
 Absorber = 25 + (331.6 / 450.3)(30 - 25) = 28.7C (decrease) 
 Condenser = 25 + (280 / 371.2)(40 - 25) = 36.3 C (decrease) 
 Evaporator = 15 - (262.4 / 348.2)(15 - 10) = 11.23 C (increase) 
 
 Take the average: 
 q
g
 = (1/2)(349.4 + 395.0) = 372.2 kW, 104 C 
 q
c
 = (1/2)(280 + 323) = 301.4 kW, 37 C 
 q
a
 = (1/2)(331.6 + 374.4) = 353 kW, 29 C 
 q
e
 = (1/2)(262.4 + 302.4) = 282.4 kW, 11 C 
 
 Full load COP
abs
 = 0.736 
 
 New COP
abs
: 
 
(increase) 0.759
372.2
282.4
q
q
COP
g
e
abs ===
 
 
18.9%or0.189
348.2
282.4-348.2
qinReduction e ==
 
 Therefore Capacity decrease by less than reduction in solution flow rate (33 1/3 %). 
Table 17-1. Influence of reduction in solution flow rate of pump
Solution concentrate Refrigerating
Component Temperature x(gen) x(abs) Capacity COP(abs)
Generator "increase" "increase"
Absorber "decrease" "increase"
Condenser "decrease"
Evaporator "increase" "decrease" "increase" 
 
 (b) Initial: 
 
( )
( )ras
asr
TTT
TTT
COP
−
−
=
 
 T
s
 = 100 C + 273 = 373 K 
 T
r
 = 10 C + 273 = 283 K 
 T
a
 = 1/2(30 C + 40 C) + 273 = 35 C + 273 = 308 K 
 
 
( )( )
( )( ) 1.973283308373
308373283
COPideal =
−
−
=
 
 COP
abs
 = 0.736 
 
 New: 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 9 of 11 
 T
s
 = 104 C + 273 = 377 K 
 T
r
 = 11 C + 273 = 284 K 
 T
a
 = 1/2(29 C + 37 C) + 273 = 33 C + 273 = 306 K 
 
 
( )( )
( )( ) (increase)2.431284306377
306377284
COPideal =
−
−
=
 
 COP
abs
 = 0.759 
 
 Then: 
 
12.683-19.1913COPCOP idealabs = 
 
( )
( ) 12.681-TTT
T-TT
19.1913COP
ras
asr
abs 





−
=
 
 (1) COP
abs
 increases as T
s
 increases: 
 
( ) ( ) 12.681-TTT
TT
TT
T
19.1913COP
ras
ra
ra
r
abs 





−
−
−
=
 
 (2) COP
abs
 increases as T
a
 decreases: 
 
( ) 12.681-T
T
TTT
)T(TT
19.1913COP
s
r
ras
rsr
abs 





−
−
−
=
 
 (3) COPabs increases as Tr increases: 
 
( )
( )
12.681-
T
T-T
TTT
)T(TT
19.1913COP
s
as
ras
asa
abs 





−
−
−
=
 
17-6. In the double-effect absorption unit shown in Fig. 17-14, LiBr-water solution leaves generator I with a 
concentration of 67 %, passes to the heat exchanger and then to generator II, where its temperature is 
elevated to 130 C. Next the solution passes through the throttling valve, where its pressure is reduced to that 
in the condenser, which is 5.62 kPa. In the process of the pressure reduction, some water vapor flashes off 
from this solution, flowing through generator II, (a) how much mass flashes to vapor. and (b) what is the 
concentration of LiBr-solution that drops into the condenser vessel? 
 
Solution: 
 At 67 %, 130 C, Fig. 17.8 
 h
1
 = -3.3 kJ/kg 
 
 At 5.62 kPa 
 
 Try t
2
 = 100 C 
 h
2
 = -55 kJ/kg solution, x2 = 68.4 % 
 h
3
 = 2676 kJ/kg water vapor 
 
 w
1
 = w
2
 + w
3
 
 w
1
x
1
 = w
2
x
2
 
 
 w
2
 / w
1
 = 0.67 / x
2
 
 
 w
1
h
1
 = w
2
h
2
 + w
3
h
3
 
 
 h
1
 = (w
2
/w
1
)h
2
 + (w
3
/w
1
)h
3
 
 -3.3 = (0.67/ x
2
)(055) + (w
3
/ w1)(2676) 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 10 of 11 
 
 w
1
 = w
2
 + w
3
 
 
 1 = (w
2
/w
1
) + (w
3
/w
1
) 
 
 1 = (0.67/x
2
) +(w
3
/w
1
) 
 
 -3.3 = (0.67/x
2
)(-55) + (1 - 0.67/x
2
)(2676) 
 
 0.67/x
2
 = 0.9811 
 
 x
2
 = 0.683 = 68.4 % 
 
 (a) Mass flashes to vapor = w3/w1 
 
 w
3
/w
1
 = 1 - (.67/x
2
) 
 
 w
3
/w
1
 = 1 - (0.67 / 0.684) 
 
 w
3
/w
1
 = 0.0205 kg/kg of solution flowing through generator II - - Ans. 
 
 
 (b) x2 = 0.684 = 68.4 % - - - Ans. 
 
17-7. The combined absorption and vapor-compression system shown in Fig. 17-16 is to be provided with a 
capacity control scheme that maintains a constant temperature of the leaving chilled water as the 
temperature of the return water to be chilled varies. This control scheme is essentially one of reducing the 
refrigerating capacity. The refrigerant compressor is equipped with inlet valves (see Chap. 11), the speed of 
the turbine-compressor can be varied so long as it remains less than the maximum value of 180 r/s, and the 
control possibilities of the absorption unit are as described in Sec. 17-11. The characteristics of the steam 
turbine are that both its speed and power diminishes if the pressure of the supply steam decreases or the 
exhaust pressure increases. With constant inlet and exhaust pressures the speed of the turbine increases if 
the load is reduced. Device a control scheme and describe the behavior of the entire system as the required 
refrigerating load decreases. 
 
Answer: 
 1. If the return water to be chilled reduces, the refrigerating capacity will be reduced. 
 2. For the refrigerating capacity reduced, the steam entering the generator of absorption unit will be 
throttled to reduce the generator temperature. 
 3. For the vapor-compression unit, the compressor can be controlled by adjusting prerotation vane at 
the impeller inlet. 
 4. For the entire system with the above control scheme, there is a possibility that the speed of turbine-
compressor will increase greater than 180 r/s. So it is better to control only the exhaust pressure by 
increasing it then throttled before entering the generator of absorption unit. The refrigerating capacity and 
power diminishes as the exhaust pressure increases with constant supply steam. 
 
17-8. The operating cost of an absorption system is to be compared with an electric-driven vapor-compression 
unit. The cost of natural gas on a heating value basis is $4.20 per gigajoule, when used as fuel in a boiler it 
has a combustion efficiency of 75 percent. An absorption unit using steam from this boiler has a COP
abs
 of 
0.73. If a vapor-compression unit is selected, the COP would be 3.4, and the electric motor efficiency is 85 
percent. At what cost of electricity are the operating costs equal? 
 
Solution: 
CHAPTER 17 - ABSORPTION REFRIGERATION 
 Page 11 of 11 
 Let q
e
 = refrigerating capacity = kWh 
 
 Operating cost of natural gas 
 = ($4.20 /GJ)(1 GJ / 106 kJ)(3600 kJ / 1 kWh)(q
e
 / 0.73)(1 / 0.75) 
 = $ 0.0276164q
e 
 
 Let x = operating cost in cents / kWh 
 
 Operating cost of electric motor. 
 = (x / 100)(q
e
) 
 
 Then: 
 (x / 100)(q
e
) = 0.0276164(q
e
) 
 
 x = 2.76 cents / kWh - - - Ans. 
 
- 0 0 0 - 
CHAPTER 18 - HEAT PUMPS 
 Page 1 of 4 
18-1. An air-source heat pump uses a compressor with the performance characteristics shown in Fig. 18-4. The 
evaporator has an air-side area of 80 m2 and a U-value of 25 W/m
2
.K. The airflow rate through the 
evaporator is 2 kg/s, and the condensing temperature is 40 C. Using the heat-rejection ratios of a hermetic 
compressor from Fig. 12-12, determine the heating capacity of the heat pump when the outdoor-air 
temperature is 0C. 
 
Solution: 
 use Fig. 18-4 and Fig. 12-2. 
 
 Fig. 12-2 at 40 C Condensing temperature. 
 Evaporating Temperature, t
e
 Heat-rejection ratio 
 10 C 1.19 
 0 C 1.255 
 -10 C 1.38 
 
 Heat-rejection ratio = 1.255 - 0.0095t
e
 + 0.0003t
e
2
 
 
 Fig. 18-4. At outdoor air temperature = 0 C 
 
 Evaporating Temperature, t
e
 Rate of evaporator heat transfer 
 -10 C 8.5 kw 
 0 C 12.9 kw 
 10 C 18.0 kw 
 
 Rate of evaporator heat transfer = 12.9 + 0.475t
e
 + 0.0035t
e
2
 
 
 For evaporator, ambient = 0 C 
 







−
−
−
=
e2
e1
21
tt
tt
ln
tt
LMTD
 
 
 q
e
 = UALMTD 
 
 At 0 C, c
pm
 = 1.02 kJ/kg.K = 1020 J/kg.K say purely sensible. 
 
 q
e
 = wc
pm
Dt = wc
pm
(t
1
 - t
2
) 
 q
e
 = (2)(1020)(0 - t2) 
 
 But, 
 
 
( )( ) ( )








−
−
−
=
e2
e
2
e
tt
t0
ln
t0
8025q
 
 Then, 
 
1.02
1
tt
t0
ln
e2
e
=







−
−
 
 
2.6655
tt
t0
e2
e
=
−
−
 
 1.6644t
e
 = 2.6644t
2
 
CHAPTER 18 - HEAT PUMPS 
 Page 2 of 4 
 t
2
 = 0.624836t
e
 
 
 q
e
 = (2)(1020)(0 - 0.624836te) / 1000 kW 
 q
e
 = -1.274665t
e
 kW 
 
 q
e
 = 12.9 + 0.475t
e
 + 0.0035t
e
2
 = -1.274665te 
 0.0035t
e
2
 + 1.749665t
e
 + 12.9 = 0 
 
 t
e
 = -7.485 C 
 
 q
e
 = -1.274665(-7.485) 
 q
e
 = 9.541 kW 
 
 Heat-rejection ratio = 1.255 - 0.0095(-7.485) + 0.0003(-7.485)
2
 
 Heat-rejection ratio = 1.343 
 
 q
c
 = (1.343)(9.541) 
 
 q
c
 = 12.8 kW - - - Ans. 
 
18-2. The heat pump and structure whose characteristics are shown in Fig. 18-6 are in a region where the deisgn 
outdoor temperature is -15 C. The compressor of the heat pump uses two cylinders to carry the base load 
and brings a third into service when needed. The third cylinder has a capacity equal to either of the other 
cylinders. How much supplementary resistance heat must be available at an outdoor temperature of -15 C? 
 
Solution: 
 Use Fig. 18-6. 
 At -15 C 
 Heat loss of structure = 17.8 kW 
 Heating capacity = 8.0 kW 
 
 For two-cylinder = 8.0 kW 
 
 For three-cylinder = (3/2)(8.0 kW) = 12.0 kW 
 
 Supplementary resistance heat = 17.8 kW - 12.0 kW 
 = 5.8 kW - - - Ans. 
 
18-3. The air-source heat pump referred to in Figs. 18-4 and 18-5 operates 2500 h during the heating season, in 
which the average outdoor temperature is 5 C. The efficiency of the compressor motor is 80 percent, the 
motor for the outdoor air fan draws 0.7 kW, and the cost of electricity is 6 cents per kilowatt-hour. What is the 
heating cost for this season. 
 
Solution: 
 Use Fig. 18-4 and Fig. 18-5. 
 Outdoor air temperature = 5 C. 
 Fig. 18-5 
 Heating capacity = 15.4 kW 
 Evaporator heat-transfer rate = 12 kW 
 Compressor Power = 3.4 kW 
 
 Power to compressor motor = (3.4)(2500)($0.06) / (0.80) 
 = $ 637.50 
CHAPTER 18 - HEAT PUMPS 
 Page 3 of 4 
 Power to outdoor air fan motor = $ 150.00 
 
 Heating Cost = $ 637.50 + $ 150.00 = $ 787.50 - - - Ans. 
 
18-4. A decentralized heat pump serves a building whose air-distribution system is divided into one interior and 
one perimeter zone. The system uses a heat rejector, water heater, and storage tank (with a water capacity 
of 60 m
3
) but no solar collector. The heat rejector comes into service when the temperature of the return-loop 
water reaches 32 C, and the boiler supplies supplementary heat when the return-loop water temperature 
drops to 15 C. Neither component operates when the loop water temperature is between 15 and 32 C. The 
heating and cooling loads of the different zones for two periods of a certain day as shown in Table 18-1. The 
loop water temperature is 15 C at the start of the day (7 A.M.). The decentralized heat pumps operate with 
COP of 3.0. Determine the magnitude of (a) the total heat rejection at the heat rejector from 7 A.M. yo 6 P.M. 
and (b) the supplementary heat provided from 6 P.M. to 7 A.M. 
 
Table 18-1 Heating and Cooling loads in Prob. 18-4. 
 Interior zone Perimeter zone 
 
 Heating, kW Cooling, kW Heating, kW Cooling, kW 
7 A.M. to 6 P.M. -------------- 260 ------------- 40 
6 P.M. to 7 A.M. -------------- 50 320 ------------ 
 
Solution: 
 
 Weight of water in storage tank. 
 V = 60 m
3
 
 at 24 C, ρ = 997.4 kg/m3 
 w = (997.4)(60) = 59,884 kg 
 
 Storage tank heat 
 = (59,884 kg)(4,190 kJ/kg.K)(32 - 15 K) 
 = 4.266 GJ 
 
(a) Heating time = 
 
( ) ( )s/h3600
3
31
kW40260
kJ4,265,537





 +
+
=
 
 = 2.962 hra 
 From 7 A.M. to 6 P.M. = 11 hrs 
 
 Total heat rejection 
 = (260 + 40 kW)[(1+ 3) / 3](3600 s/h)(11 - 2.962 hr) 
 = 11,574,720 kJ 
 = 11.6 GJ - - - Ans. 
 
 (b) Supplementary heat 
 
 Storage tank heat = 4,265,537 kJ 
 
 
( ) kJ4,265,537s/h3600kw
3
31
50
31
3
320(Time) =










 +
−





+
 
 Time = 6.8358 hrs 
 
 Supplementary heat 
CHAPTER 18 - HEAT PUMPS 
 Page 4 of 4 
 
( )( )hr6.835813s/h3600kw
3
31
50
31
3
320 −










 +
−





+
=
 
 = 3,846,461 kJ 
 = 3.85 GJ - - - Ans. 
 
18-5. The internal-source heat pump using the double-bundle heat pump shown in Fig. 18-9 is to satisfy a heating 
load of 335 kW when the outdoor temperature is -5 C, the return air temperature is 21 C, and the 
temperature of the cool supply air is 13 C. The minimum percentage of outdoor air specified for ventilation is 
15 percent, and the flow rate of cool supply air is 40 kg/s. If the COP of the heat pump at this condition is 3.2, 
how much power must be provided by the supplementary heater? 
 
Solution: 
 Outdoor air = -5 C, 15 % flow rate 
 Return air = 21 C 
 t
3
 = Cool supply air = 13 C, w = 40 kg/s 
 COP = 3.2 
 Heating Load = 335 kW 
 
 t
4
 = mix temperature = (0.15)(-5 C) + (0.85)(21 C) 
 t
4
 = 17.1 C 
 
 q
e
 = wc
p
(t
4
 - t
3
) 
 q
e
 = (40 kg/s)(1.0 kJ/kg.K)(17.1 C - 13 C) 
 q
e
 = 164 kW 
 
 Condenser 
 q
c
 = q
e
(1+ COP) / COP 
 q
c
 = 164(1 + 3.2) / 3.2 
 q
c
 = 215.25 kW 
 
 Supplementary heat = 335 kW - 215.25 kW 
 = 119.75 kW - - - Ans. 
 
 
- 0 0 0 - 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 1 of 10 
19-1. Another rating point from the cooling tower catalog from which the data in Example 19-1 are taken specifies 
a reduction in water temperature from 33 to 27 C when the entering-air enthalpy is 61.6 kJ/kg. The water flow 
rate is 18.8 kg/s, and the air flow rate is 15.6 kg/s. Using a stepwise integration with 0.5-K increments of 
change in water temperature, compute h
c
A/c
pm
 for the tower. 
 
 
Solution: 
 Eq. 19-4. 
 
( )∑ −∆=
maipm
c
hh
1
t4.19L
c
Ah
 
 L = 18.8 kg/s 
 G = 15.6 kg/s 
 t
 in
 = 33 C 
 t 
out
 = 27 C 
 
 Use 12-section, 0.5 K water drop in each section. 
 Eq. 19-1 
 dq = gdh
a
 = L (4.19 kJ/kg.K)dt kW 
 
 Entering air enthalpy = 61.6 kJ/kg 
 
 For section 0-1, 27 to 27.5 C 
 
( )K0.54.19
G
L
hh a,0a,1 =−
 
 h
a,0
 = 61.6 kJ/kg 
 
 
( ) 2.53K0.54.19
15.6
18.8
61.6ha,1 =





=−
 
 h
a,1
 = 64.13 kJ/kg 
 
 Average h
a
 = (1/2)(h
a,0
 + h
a,1
) 
 = (1/2)(61.6 + 64.13) 
 = 62.86 kJ/kg 
 
 Mean water temperature = 27.25 C 
 From Table A-2, 
 Average h
i
 = 86.44 kJ/kg 
 
 (h
i
 - h
a
)
m
 = 86.44 kJ/kg - 62.86 kj/kg 
 = 23.58 kJ/kg 
 
Table 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 2 of 10 
Section Mean Water Av erage ha, Av erage hi, (hi-ha)m 1/(hi-ha)m
Temp., C kJ/kg kJ/kg kJ/kg
0-1 27.25 62.86 86.44 23.58 0.04241 
1-2 27.75 65.39 88.78 23.39 0.04275 
2-3 28.25 67.92 91.18 23.26 0.043 
3-4 28.75 70.45 93.63 23.18 0.04314 
4-5 29.25 72.98 96.13 23.15 0.0432 
5-6 29.75 75.51 98.7 23.19 0.04312 
6-7 30.25 78.04 101.32 23.28 0.04296 
7-8 30.75 80.57 104 23.43 0.04269 
8-9 31.25 83.1 106.74 23.64 0.0423 
9-10 31.75 85.63 109.54 23.91 0.04182 
10-11 32.25 88.16 112.41 24.25 0.0412411-12 32.75 90.69 115.35 24.66 0.04055 
 
 
( ) 0.50917hh
1
mai
=
−
∑
 
Eq. 19-4. 
( )( )( )0.509170.518.84.19
c
Ah
pm
c
=
 
= 20.0 kW/(kJ/kg of enthalpy difference) - - - Ans. 
 
19-2. Solve Prob. 19-1 using a compute program and 0.1-K increments of change of water temperature. 
 
Solution: 
 
 Formula: n = 0 to 60 
 mean water temperature = (1/2)(t
o
 + t
i
) 
 or = (1/2)(
t
n
 + t
n+1
) - - Eq. 1 
 
 Mean air enthalpy 
 h
a,1
 - h
a,0
 = (L/G)(4.19)(0.1 K) 
 = (18.0 / 15.6)(4.19)(0.1) 
 = 7.542 / 15.6 
 
 h
a,0
 = 61.6 kJ/kg 
 
 h
a,1
 = h
a,0
 + 7.542/15.6 
 
 h
a
 = h
a,o
 + 3.771/15.6 
 
 h
a
 = h
a,n
 + 3.771/16.5 - - Eq. 2 
 
 Mean h
i
 
 Equation 19-5 
 
 h
i
 = 4.7926 + 2.568t - 0.029834t
2
 + 0.0016657t
3
 - - Eq. 3 
 
 where t = mean water temperature 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 3 of 10 
 
 (h
i
 - h
a
)m = mean h
i
 - mean air enthalpy - - - Eq. 4 
 
Program; Microsoft Spreadsheet 
 
Row 1: 
A1 = “Water Temp. C” 
C1 = “Mean Water Temp., C” 
E1 = “Mean Air Enthalpy, kJ/kg” 
F1 = “(hi - ha)m” 
G1 = “1/(hi - ha)m” 
Row 2: 
A2 = “tn” 
B2 = “tn+1” 
Row 3: 
A3 = 27 
B3 = A3 + 0.1 
C3 = (1/2)(A3 + B3) 
D3 = 61.6 + 3.771/15.6 
E3 = 4.7926 + 2.568*C3 - 0.029834*C3^2 + 0.0016657*C3^3 
F3 = E3 - D3 
G3 = 1/F3 
Row 4: 
A4 = B3 
B4 = A4 + 0.1 
C4 = (1/2)(A4 + B3) 
D4 = D3 + 7.542/15.6 
E4 = 4.7926 + 2.568*C4 - 0.029834*C4^2 + 0.0016657*C4^3 
F4 = E4 - D4 
G4 = 1/F4 
Row 5: 
A5 = B4 
B5 = A5 + 0.1 
C5 = (1/2)(A5 + B5) 
D5 = D4 + 7.542/15.6 
E5 = 4.7926 + 2.568*C5 - 0.029834*C5^2 + 0.0016657*C5^3 
F5 = E5 - D5 
G5 = 1/F5 
 
up to 62 rows 
Row 62: 
A62 = B61 
B62 = A62 + 0.1 
C62 = (1/2)(A62 + B62) 
D62 = D61 + 7.542/15.6 
E62 = 4.7926 + 2.568*C62 - 0.029834*C62^2 + 0.0016657*C62^3 
F62 = E62 - D62 
G62 = 1/F62 
 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 4 of 10 
Water Temp. 
C
Mean 
Water 
Temp. 
C
Mean Air 
Enthalpy, 
kJ/kg
Mean hi, 
kJ/kg
(hi - ha)m 1/(hi - ha)m
tn tn+1
27 27.1 27.05 61.841731 85.395843 23.554112 0.0424554 
27.1 27.2 27.15 62.325192 85.857935 23.532742 0.042494 
27.2 27.3 27.25 62.808654 86.322144 23.51349 0.0425288 
27.3 27.4 27.35 63.292115 86.788479 23.496364 0.0425598 
27.4 27.5 27.45 63.775577 87.256952 23.481375 0.0425869 
27.5 27.6 27.55 64.259038 87.727571 23.468532 0.0426102 
27.6 27.7 27.65 64.7425 88.200347 23.457847 0.0426297 
27.7 27.8 27.75 65.225962 88.675289 23.449328 0.0426451 
27.8 27.9 27.85 65.709423 89.152408 23.442985 0.0426567 
27.9 28 27.95 66.192885 89.631714 23.43883 0.0426642 
28 28.1 28.05 66.676346 90.113217 23.436871 0.0426678 
28.1 28.2 28.15 67.159808 90.596926 23.437119 0.0426674 
28.2 28.3 28.25 67.643269 91.082852 23.439583 0.0426629 
28.3 28.4 28.35 68.126731 91.571005 23.444274 0.0426543 
28.4 28.5 28.45 68.610192 92.061394 23.451202 0.0426417 
28.5 28.6 28.55 69.093654 92.554031 23.460377 0.0426251 
28.6 28.7 28.65 69.577115 93.048923 23.471808 0.0426043 
28.7 28.8 28.75 70.060577 93.546083 23.485506 0.0425795 
28.8 28.9 28.85 70.544038 94.045519 23.50148 0.0425505 
28.9 29 28.95 71.0275 94.547241 23.519741 0.0425175 
29 29.1 29.05 71.510962 95.051261 23.540299 0.0424803 
29.1 29.2 29.15 71.994423 95.557587 23.563164 0.0424391 
29.2 29.3 29.25 72.477885 96.066229 23.588345 0.0423938 
29.3 29.4 29.35 72.961346 96.577198 23.615852 0.0423444 
29.4 29.5 29.45 73.444808 97.090504 23.645697 0.042291 
29.5 29.6 29.55 73.928269 97.606157 23.677887 0.0422335 
29.6 29.7 29.65 74.411731 98.124166 23.712435 0.042172 
29.7 29.8 29.75 74.895192 98.644541 23.749349 0.0421064 
29.8 29.9 29.85 75.378654 99.167294 23.78864 0.0420369 
29.9 30 29.95 75.862115 99.692432 23.830317 0.0419634 
30 30.1 30.05 76.345577 100.21997 23.874391 0.0418859 
30.1 30.2 30.15 76.829038 100.74991 23.920871 0.0418045 
30.2 30.3 30.25 77.3125 101.28227 23.969768 0.0417192 
30.3 30.4 30.35 77.795962 101.81705 24.021092 0.0416301 
30.4 30.5 30.45 78.279423 102.35428 24.074852 0.0415371 
30.5 30.6 30.55 78.762885 102.89394 24.131059 0.0414404 
30.6 30.7 30.65 79.246346 103.43607 24.189722 0.0413399 
30.7 30.8 30.75 79.729808 103.98066 24.250852 0.0412357 
30.8 30.9 30.85 80.213269 104.52773 24.314458 0.0411278 
30.9 31 30.95 80.696731 105.07728 24.380551 0.0410163 
31 31.1 31.05 81.180192 105.62933 24.449141 0.0409012 
31.1 31.2 31.15 81.663654 106.18389 24.520236 0.0407826 
31.2 31.3 31.25 82.147115 106.74096 24.593849 0.0406606 
31.3 31.4 31.35 82.630577 107.30056 24.669988 0.0405351 
31.4 31.5 31.45 83.114038 107.8627 24.748663 0.0404062 
31.5 31.6 31.55 83.5975 108.42739 24.829885 0.040274 
31.6 31.7 31.65 84.080962 108.99463 24.913664 0.0401386 
31.7 31.8 31.75 84.564423 109.56443 25.000008 0.04 
31.8 31.9 31.85 85.047885 110.13681 25.08893 0.0398582 
31.9 32 31.95 85.531346 110.71178 25.180438 0.0397134 
32 32.1 32.05 86.014808 111.28935 25.274542 0.0395655 
32.1 32.2 32.15 86.498269 111.86952 25.371253 0.0394147 
32.2 32.3 32.25 86.981731 112.45231 25.47058 0.039261 
32.3 32.4 32.35 87.465192 113.03773 25.572534 0.0391045 
32.4 32.5 32.45 87.948654 113.62578 25.677124 0.0389452 
32.5 32.6 32.55 88.432115 114.21648 25.78436 0.0387832 
32.6 32.7 32.65 88.915577 114.80983 25.894253 0.0386186 
32.7 32.8 32.75 89.399038 115.40585 26.006813 0.0384515 
32.8 32.9 32.85 89.8825 116.00455 26.122049 0.0382818 
32.9 33 32.95 90.365962 116.60593 26.239971 0.0381098 
SUM = 2.4810051 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 5 of 10 
( ) 2.481hh
1
mai
=
−
∑
 
Eq. 19-4. 
( )( )( )2.4810.118.84.19
c
Ah
pm
c
=
 
= 19.54 kW/(kJ/kg of enthalpy difference) - - - Ans. 
 
19-3. If air enters the cooling tower in Prob. 19-1 with a dry-bulb temperature of 32 C, compute the dry-bulb 
temperatures as the air passes thorough the tower. For the stepwise calculation choose a change in water 
temperature of 0.5 K, for which the values of 1/(h
i
-h
a
)
m
 starting at the bottom section are, respecitively, 
0.04241, 0.04274, 0.04299, 0.04314, 0.04320, 0.04312, 0.04296, 0.04268, 0.04230, 0.04182, 0.04124, and 
0.04055. 
 
Solution: t
a,0
 = 32 C 
 
For section 0-1 
 
( ) 0.04241hh
1
mai
=
−
 
Dividing Eq. 19-7 by 2G. 
 
( )( )( )( )
( )( ) 0.0535415.62
0.042410.518.84.19
2Gc
Ah
pm
c
==
∆
 
From Eq. 19-6. 
( )( )
C31.36
0.053541
27.52735.00.0535432.0
t a,1 =
+
−−−
=
 
 
Tabulation: 
n section 
( )
mai
hh
1
−
 
pm
c
2Gc
Ah ∆
 t
a.n+1 
0 0-1 0.04241 0.05354 31.36 
1 1-2 0.04274 0.05395 30.99 
2 2-3 0.04299 0.05427 30.71 
3 3-4 0.04314 0.05446 30.51 
4 4-5 0.04320 0.05453 30.38 
5 5-6 0.04312 0.05443 30.31 
6 6-7 0.04296 0.05423 30.30 
7 7-8 0.04268 0.05388 30.35 
8 8-9 0.04230 0.05340 30.44 
9 9-10 0.04182 0.05279 30.57 
10 10-11 0.04124 0.05206 30.74 
11 11-12 0.04055 0.05119 30.94 
 
 t
a,12
 = 30.94 C - - - Ans. 
 
19-4. A crossflow cooling tower operating with a water flow rate of 45 kg/s and an airflow rate of 40 kg/s has a 
value of h
c
A/c
pm
 = 48 kW/(kJ/kg of enthalpy difference). The enthalpy of the entering air is 80 kJ/kg, and the 
temperature of entering water is 36 C. Develop a computer program to predict the outlet water temperature 
when the tower is divided into 12 sections, as illustrated in Fig. 19-8. 
 
Solution: Refer to Fig. 19-8. 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 6 of 10 
 
 Water flow rate = 45 kg/s 
 Air flow rate = 40 kg/s 
 
 h
c
A/c
pm
 = 48 kW/(kJ/kg of enthalpy difference 
 
 t
in
 = 36 C 
 h
in
 = 80 kJ/kgTable A-2 at 36 C 
 h
i, in
 = 136.16 kJ/kg 
 
 For section 1. 
 L = 45 / 4 = 11.25 kg/s 
 G = 40 / 3 = 13.33 kg/s = 40/3 kg/s 
 
 (h
c
A/c
pm
)/12 = 48/12 = 4.0 kW/(kJ/kg of enthalpy difference 
 
 Eq. 19-8. 
 q = (11.25)(4.19)(t
in
 - t
1
) 
 Eq. 19-9. 
 q = (40 / 3)(h
1
 - h
in
) 
 Eq. 19-10. 
 







 +
−
−






=
2
hh
2
hh
kJ/kg
kW
4q 1in
outi,ini,
 
 Combination of Eq. 19-9 and Eq. 19-10. 
 
( )[ ]( )
( ) G2cAh
hth136.162cAhGh
h
pmc
inoui,pmcin
1
+∆
−+∆+
=
 
 Eq. 19-5 
 h
i,out
 = 4.7926 + 2.568t
1
 - 0.029834t
1
2
 + 0.0016657t
1
3
 
 
For section 1, 
1. Eq. 19-5. 
2. h
in
 = 80 kJ/kg, G = 40/3 kg/s 
3. Combination of Eq. 19-0 and Eq. 19-10. 
4. q = (40 / 3)(h
1
 - h
in
) 
5. t
in
 = 36 C 
6. q = (11.25)(4.19)(t
in
 - t
1
) solve for t
1
 
 
Then. ( )[ ]( )
( ) G2cAh
hth136.162cAhGh
h
pmc
inoui,pmcin
1
+∆
−+∆+
=
 
h
1
 = 76.8904 + 0.13044h
i, out
 
 
q = (40 / 3)(76.8904 + 0.13044h
i,out
 - 80) 
 
q = -41.4613 + 1.7392h
i,out 
 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 7 of 10 
( )( )4.1911.25
1.7392h41.4613
36t
outi,
1
+−
−=
 
t
1
 = 36.8796 - 0.03690h
i, out
 
t
1
 = 36.8796 - 0.03690 (4.7926 + 2.568t
1
 - 0.029834t
1
2
 + 0.0016657t
1
3
) 
0.000061464t
1
3
 - 0.001109t
1
2
 + 1.09476t
1
 - 36.7028 = 0 
 
Try t
1
 = 32 C 
 f(t
1
) = 0.000061464t
1
3
 - 0.001109t
1
2
 + 1.09476t
1
 - 36.7028 = -0.7920 < 0.0000 
 
 Try t
1
 = 33 C 
 f(t
1
) = 0.4254 > 0.0000 
 
 Try t
1
 = 32.5 C, f(t
1
) = -0.1845 < 0.0000 
 Try t
1
 = 32.6 C, f(t
1
) = -0.0628 < 0.0000 
 Try t
1
 = 32.7 C, f(t
1
) = 0.0591 > 0.0000 
 Try t
1
 = 32.65 C, f(t
1
) = 0.0000 = 0.0000 
 
 Then t
1
= 32.65 C 
For computer program (Spreadsheet) 
 
Table Data: 
1. Section No. 
2. Entering Water Temperature 
3. Entering Air Enthalpy 
4. Entering Enthalpy of Saturated Air 
5. Leaving Water Temperature (Trial Value) 
6. Leaving Air Enthalpy 
7. Leaving Enthalpy of Saturated Air 
8. Leaving Water Temperature (Actual Value) 
 
Formula: 
 
Section 1 
Entering water temperature = t
in
 
Entering Air Enthalpy = h
in
 
 ( )[ ]( )
( ) G2cAh
hthh2cAhGh
h
pmc
inoui,ini,pmcin
1
+∆
−+∆+
=
 ( ))-h hGq in1= ( )[ ]( )
( ) 





−
+∆
−+∆+
= in
pmc
inoui,ini,pmcin
h
G2cAh
hthh2cAhGh
Gq
 
4.19L
q
tt in1 −=
 
( )( )
( )[ ]( )
( )( ) 





+∆
−−+∆+
+∆
−=
inpmc
inouti,ini,pmcin
pmc
in1
hG2cAh
hhh2cAhGh
G2cAh4.19L
G
tt
 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 8 of 10 
( )( )
( )( ) ( )inouti,ini,pmc
pmc
in1 hhh
G2cAh4.19L
2cAhG
tt 2−+
+∆
∆
−=
 
Eq. 19-5 
3
in
2
ininini, 0.0016657t0.029834t2.568t4.7926h +−+= 
3
out
2
outoutouti, 0.0016657t0.029834t2.568t4.7926h +−+= 
 
Entering Values: 
( )
( )inouti,ini,in1 2hhh
3
40
2
4
11.254.19
2
4
3
40
tt −+






+












−=
 
( )inouti,ini,in1 2hhh
2168.325
80
tt −+





−=
 
 
Subscript “in” is replaced in any section by subscript of its entering conditions. 
Subscript “1” is replaced in any section by subscript of its leaving conditions. 
 
Programming by spreadsheet: 
Note. 
1. Trial value should equal actual leaving water temperature in the Table by trial and error. 
 2. For sections 1, 2, 3 and 4, t
in
 = 36 C. 
 3. For section 1,5 and 9, h
in
 = 80 kJ/kg. 
 
PROGRAM: 
Row 1 
A1 = “Section No.” 
B1 = “Entering Water Temp., C” 
C1 = “Entering Air Enthalpy, kJ/kg” 
D1 = “Entering Enthalpy of Saturated Air, kJ/kg” 
E1 = ”Leaving WaterTemp.,C (Trial)” 
F1 = “Leaving Air Enthalpy, kJ/kg” 
G1 = “Leaving Enthalpy of Saturated Air, kJ/lg” 
H1 = “Leaving Water Temp., C (Actual)” 
Row 2 
A2 = 1 
B2 = 36 
C2 = 80 
D2 = 4.7926 + 2.568*B2 - 0.029834*B2^2 + 0.0016657*B2^3 
E2 = INPUT (Trial Value) 
G2 = 4.7926 + 2.568*E2 - 0.029834*E2^2 + 0.0016657*E2^3 
H2 = B2 - (80/2168.325)(G2 + D2 - 2*C2) 
Row 3 
A3 = A2 + 1 
B3 = B2 
C3 = F2 
D3 = 4.7926 + 2.568*B3 - 0.029834*B3^2 + 0.0016657*B3^3 
E3 = INPUT (Trial Value) 
G3 = 4.7926 + 2.568*E3 - 0.029834*E3^2 + 0.0016657*E3^3 
H3 = B3 - (80/2168.325)(G3 + D3 - 2*C3) 
Row 4 
A4 = A3 + 1 
B4 = B2 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 9 of 10 
C4 = F3 
D4 = 4.7926 + 2.568*B4 - 0.029834*B4^2 + 0.0016657*B4^3 
E4 = INPUT (Trial Value) 
G4 = 4.7926 + 2.568*E4 - 0.029834*E4^2 + 0.0016657*E4^3 
H4 = B4 - (80/2168.325)(G4 + D4 - 2*C4) 
Row 5 
A5 = A4 + 1 
B5 = B2 
C5 = F4 
D5 = 4.7926 + 2.568*B5 - 0.029834*B5^2 + 0.0016657*B5^3 
E5 = INPUT (Trial Value) 
G5 = 4.7926 + 2.568*E5 - 0.029834*E5^2 + 0.0016657*E5^3 
H5 = B5 - (80/2168.325)(G5 + D5 - 2*C5) 
Row 6 
A6 = A5 + 1 
B6 = H2 
C6 = 80 
D6 = 4.7926 + 2.568*B6 - 0.029834*B6^2 + 0.0016657*B6^3 
E6 = INPUT (Trial Value) 
G6 = 4.7926 + 2.568*E6 - 0.029834*E6^2 + 0.0016657*E6^3 
H6 = B6 - (80/2168.325)(G6 + D6 - 2*C6) 
Row 7 
A7 = A6 + 1 
B7 = H3 
C7 = F6 
D7 = 4.7926 + 2.568*B7 - 0.029834*B7^2 + 0.0016657*B7^3 
E7 = INPUT (Trial Value) 
G7 = 4.7926 + 2.568*E7 - 0.029834*E7^2 + 0.0016657*E7^3 
H7 = B7 - (80/2168.325)(G7 + D7 - 2*C7) 
Row 8 
A8 = A7 + 1 
B8 = H4 
C8 = F7 
D8 = 4.7926 + 2.568*B8 - 0.029834*B8^2 + 0.0016657*B8^3 
E8 = INPUT (Trial Value) 
G8 = 4.7926 + 2.568*E8 - 0.029834*E8^2 + 0.0016657*E8^3 
H8 = B8 - (80/2168.325)(G8 + D8 - 2*C8) 
Row 9 
A9 = A8 + 1 
B9 = H5 
C9 = F8 
D9 = 4.7926 + 2.568*B9 - 0.029834*B9^2 + 0.0016657*B9^3 
E9 = INPUT (Trial Value) 
G9 = 4.7926 + 2.568*E9 - 0.029834*E9^2 + 0.0016657*E9^3 
H9 = B9 - (80/2168.325)(G9 + D9 - 2*C9) 
Row 10 
A10 = A9 + 1 
B10 = H6 
C10 = 80 
D10 = 4.7926 + 2.568*B10 - 0.029834*B10^2 + 0.0016657*B10^3 
E10 = INPUT (Trial Value) 
G10 = 4.7926 + 2.568*E10 - 0.029834*E10^2 + 0.0016657*E10^3 
H10 = B10 - (80/2168.325)(G10 + D10 - 2*C10) 
Row 11 
A11 = A10 + 1 
B11 = H7 
CHAPTER 19 - COOLING TOWERS AND EVAPORATIVE CONDENSERS 
 
 Page 10 of 10 
C11 = F10 
D11 = 4.7926 + 2.568*B11 - 0.029834*B11^2 + 0.0016657*B11^3 
E11 = INPUT (Trial Value) 
G11 = 4.7926 + 2.568*E11 - 0.029834*E11^2 + 0.0016657*E11^3 
H11 = B11 - (80/2168.325)(G11 + D11 - 2*C11) 
Row 12 
A12 = A11 + 1 
B12 = H8 
C12 = F11 
D12 = 4.7926 + 2.568*B12 - 0.029834*B12^2 + 0.0016657*B12^3 
E12 = INPUT (Trial Value) 
G12 = 4.7926 + 2.568*E12 - 0.029834*E12^2 + 0.0016657*E12^3 
H12 = B12 - (80/2168.325)(G12 + D12 - 2*C12) 
Row 13 
A13 = A12 + 1 
B13 = H9 
C13 = F12 
D13 = 4.7926 + 2.568*B13 - 0.029834*B13^2 + 0.0016657*B13^3 
E13 = INPUT (Trial Value) 
G13 = 4.7926 + 2.568*E13 - 0.029834*E13^2 + 0.0016657*E13^3 
H13 = B13 - (80/2168.325)(G13 + D13 - 2*C13) 
 
Output: 
Section 
No.
Entering 
Water 
Temp., C
Entering Air 
Enthalpy, 
kJ/kg
Entering 
Enthalpy of 
Saturated 
Air, kJ/kg
Leaving 
Water 
Temp., C 
(Trial)
Laeaving 
Air 
Enthalpy, 
kJ/kg
Leaving 
Enthalpy of 
Saturated 
Air, kJ/kg
Leaving 
Water 
Temp., C 
(Actual)
1 36.0000 80.0000 136.2906 32.6409 91.8755 114.7557 32.6409 
2 36.0000 91.8755 136.2906 33.3574 101.2179 119.0839 33.3575 
3 36.0000 101.2179 136.2906 33.9182 108.5777 122.5695 33.9182 
4 36.0000 108.5777 136.2906 34.3581 114.3823 125.3649 34.3582 
5 32.6409 80.0000 114.7555 30.5199 87.4983 102.7312 30.5199 
6 33.3575 87.4983 119.0843 31.4423 94.2689 107.8193 31.4424 
7 33.9182 94.2689 122.5697 32.2114100.3031 112.2270 32.2115 
8 34.3582 100.3031 125.3654 32.8534 105.6230 116.0250 32.8534 
9 30.5199 80.0000 102.7312 29.114 84.9703 95.3750 29.1140 
10 31.4424 84.9703 107.8196 30.0389 89.9319 100.1613 30.0389 
11 32.2115 89.9319 112.2277 30.8503 94.7443 104.5294 30.8503 
12 32.8534 94.7443 116.0252 31.5611 99.3131 108.4902 31.5611 
 
- 0 0 0 - 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 1 of 12 
20-1. Using Eq. 20-3, compute the hour of sunrise on the shortest day of the year of 40
o
 north latitude. 
 
Solution: Eq. 20-3 
 
 sin β = cos L cos H cos δ + sin L sin δ 
 H = hour angle = 15 degrees per one hour from solar noon. 
 L = 40
o
 
 
 β = 0o - solar altitude at sunrise 
 δ = -23.5o - shortest day on winter solstice 
 
 sin 0 = cos 40 cos H cos (-23.5) + sin 40 sin (-23.5) 
 
 H = 68.6
o
 
 
 or 68.6 / 15 = 4.573333 hrs from noon 
 = 12 - 4.573333 
 = 7.426667 from midnight 
 = 7:26 A.M. - - - Ans. 
 
20-2. Compute the solar azimuth angle at 32
o
 north latitude on February 21. 
 
Solution: From Table 4-13 
 Solar Time A.M. β φ 
 7 7 73 
 8 18 64 
 9 29 53 
 10 38 39 
 11 45 21 
 12 47 0 
 
β = solar altitude 
φ = solar azimuth 
 
Eq. 20-4 
β
δ
=φ
cos
sinHcos
sin
 
o90for ≤φ
 
Eq. 20-3. 
sin β = cos L cos H cos δ + sin L sin δ 
 
L = latitude = 32
o
 
H = Hour Angle 
δ = Declination angle 
 
For February 21 
N = 31 + 21 = 52 
 
Eq. 20-2. 
( )
365
N284360
23.47sin
+
=δ
 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 2 of 12 
( )
365
284360
23.47sin
52+
=δ
 
δ = -11.24o 
sin β = cos L cos H cos δ + sin L sin δ 
sin β = cos 32 cos H cos (-11.24) + sin 32 sin (-11.24) 
sin β = 0.83178 cos H - 0.10329 
β
δ
=φ
cos
sinHcos
sin
 ( )
β=φ cos
sinH-11.24cos
sin
 






β=φ cos
H0.98082sin
Arcsin
 
Then: H = 15
o
 x (No. of hours from Noon) 
 
Ans. 
Tabulation 
Solar Time, A.M. H β φ 
 7 75 6.43 72.44 
 8 60 18.22 63.41 
 9 45 29.00 52.46 
 10 30 38.10 38.55 
 11 15 44.44 20.83 
 12 0 46.76 0.00 
 
20-3. (a) What is the angle of incidence of the sun’s rays with a south-facing roof that is sloped at 45
o
 with the 
horizontal at 8 A.M. on June 21 at a latitude of 40
o
 north? (b) What is the compass direction of the 
sun at this time? 
 
Solution: 
 S = 45
o
 
 L = 40
o
 
 At 8 A.M. 
 H = 4 x 15 = 60
o
 
 On June 21, δ = 23.5o 
 
 (a) Eq. 20-3. 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 60 cos 23.5 + sin 40 sin 23.5 
 β = 37.41 
 β
δ
=φ
cos
sinHcos
sin
 
 37.41 cos
60 sin 23.5 cos
sin =φ
 
 φ = 30.83 
 Table 4-13, φ > 90 
 φ = 180 - 89.04 = 90.96 
 
ϕ±φ=γ
 
 
0=ϕ
 
 γ = 90.96- 0 = 90.96 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 3 of 12 
 Eq. 20-8. 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 cos θ = cos 37.41 cos 90.96 sin 45 + sin 37.41 cos 45 
 θθθθ = 65o - - - Ans. 
 
 (b) Compass direction = 65
o
 SE - - - - Ans. 
 
20-4. As an approach to selecting the tilt angle Σ of a solar collector a designer chooses the sum of I
DN
cosθ at 10 
A.M. and 12 noon on January 21 as the criterion on which to optimize the angle. At 40
o
 north latitude, with 
values of A = 1230 W/m
2
 and B = 0.14 in Eq. (20-9), what is the optimum tilt angle? 
 
Solution: 
 Eq. 20-9. 
 
( )β= sinBexp
A
IDN
 
 A = 1230 W/m
2
 
 B = 0.14 
 L = 40
o
 
 
 At 10 A.M. January 21. 
 
( )
365
N284360
23.47sin
+
=δ
 
 N = 21 
 
( )
20.16
365
21284360
23.47sin −=
+
=δ
 
 H = 2 x 15 = 30 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 30 cos (-20.16) + sin 40 sin (-20.16) 
 β = 23.66 
 β
δ
=φ
cos
sinHcos
sin
 
 
 
 
( )
cos23.66
sin30-20.16cos
sin =φ
 
 φ = 30.83 
 
 At 12 NOON January 21. 
 
( )
365
N284360
23.47sin
+
=δ
 
 N = 21 
 
( )
20.16
365
21284360
23.47sin −=
+
=δ
 
 H = 0 x 15 = 0 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 0 cos (-20.16) + sin 40 sin (-20.16) 
 β = 29.84 
 β
δ
=φ
cos
sinHcos
sin
 
 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 4 of 12 
 
 
( )
cos29.84
0sin-20.16cos
sin =φ
 
 φ = 0.0 
Then, 
At 10 A.M., β = 23.66, φ = 30.83 
At 12 N.N., β = 29.84, φ = 0.0 
Eq. 20-8. 
 
cos θ = cos β cos γ sin Σ + sin β cos Σ 
 
Then cos γ = cosφ 
 
Subsitute in Eq. 20-9. 
 
At 10 A.M. 
 
( )
( )β
Σβ+Σγβ
=θ
sinBexp
cossinsincoscosA
cosIDN
 
 
( ) ( ) ( ) ( )( )
( )( )23.66sin0.14exp
cos23.66sinsin30.83cos23.66cos1230
cosIDN
Σ+Σ
=θ
 
 
( )( )
1.417449
sΣ0.401308conΣ0.786513si1230
cosIDN
+
=θ
 
 
Σ348.238cosΣ682.502sincosIDN +=θ 
 
At 12 NN. 
 
( )
( )β
Σβ+Σγβ
=θ
sinBexp
cossinsincoscosA
cosIDN
 
 
( ) ( ) ( ) ( )( )
( )( )29.84sin0.14exp
cos29.84sinsin0.0cos29.84cos1230
cosIDN
Σ+Σ
=θ
 
 
( )( )
1.324933
sΣ0.497580conΣ0.867418si1230
cosIDN
+
=θ
 
 
Σ461.928cos05.266sinΣcosIDN +=θ 8 
Total: 
( ) ( ) Σ++Σ+= cos461.928348.238sin805.26682.502T
 
Σ+Σ= 810.166cos1487.77sinT 
Differentiate then equate to zero. 
0810.166sin1487.77cosT =Σ−Σ=′ 
810.166
1487.77
tan =Σ
 
ΣΣΣΣ = 61.43o - - - - Ans. 
 
20-5. Plot the efficiency of the collector described in Example 20-3 versus temperature of fluid entering the 
absorber over the range of 30 to 140 C fluid temperatures. The ambient temperature is 10 C. If the collector 
is being irradiated at 750 W/m
2
, determine the rate of collection at entering fluid temperatures at (a) 50 C and 
(b) 100 C. 
 
Solution: Refer to Example 20-3. 
 
10Ct =
∞ 
 t
ai
= 30 to 140 C 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 5 of 12 
 I
iθ = 800 W/m2 
 Fr = 0.90 
 α
a
 = 0.90 
 
0.87ττ c2c1 == 
 
 Eq. 20-12. 
 
( )
Fr
I
Utt
I
A
q
i
ai
c2c1
i
a







 −
−αττ==η
θ
∞
θ
a
 
 
( )( )( ) ( ) ( )0.9
800
3.510t
0.90.870.87 ai 




 −
−=η
 
 
Tabulation: 
 t
ai
 η 
 30 0.534 
 40 0.495 
 50 0.456 
 60 0.416 
 70 0.377 
 80 0.338 
 90 0.298 
 100 0.259 
 110 0.219 
 120 0.180 
 130 0.140 
 140 0.101 
Plot: 
 
 
(a) At 50 C, I
iθ = 750 W/m
2
 
( )( )FrttUI
A
q
aic2c1i
a
∞θ −−αττ= a
 
 
( )( )( )( ) ( )( )[ ]( )0.910503.50.90.870.87750
A
qa
−−=
 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 6 of 12 
q
a
/A = 334 W/m
2
 - - - Ans. 
 
(a) At 100 C, I
iθ = 750 W/m
2
 
( )( )FrttUI
A
q
aic2c1i
a
∞θ −−αττ= a
 
 
( )( )( )( ) ( )( )[ ]( )0.9101003.50.90.870.87750
A
qa
−−=
 
q
a
/A = 176 W/m
2
 - - - Ans. 
 
20-6. A 1.25- by 2.5-m flat-plate collector receives solar irradiation at a rate of 900 W/m2. It has a single cover 
plate with τ = 0.9, and the absorber has an absorptivity of α
a
 = 0.9. Experimentally determined values are Fr 
= 0.9 and U = 6.5 W/m
2
.K. The cooling fluid is water. If the ambient temperature is 32 C and the fluid 
temperature is 60 C entering the absorber, what are (a) the collector efficiency, (b) the fluid outlet 
temperature for a flow rate of 25 kg/h, and (c) the inlet temperature to the absorber at which output drops to 
zero? 
 
Solution: 
 (a) Eq. 20-12. 
 
( )
Fr
I
Utt
i
ai
c2c1 






 −
−αττ=η
θ
∞
a
 
 
 
 
0.90ττ c2c1 == 
 α
a
 = 0.90 
 Fr = 0.90 
 U = 6.5 W/m
2
.K 
 
( )( ) ( ) ( )0.9
900
3.532
0.90.9 




 −
−=η 60
 
 ηηηη = 0.701 - - - Ans. 
 
 (b) 
 
θ
=η
i
a
IA
q
 
 A = 1.25 x 2.5 = 3.125 m
2
 
 
 I
iθ = 900 W/m2 
 η = 0.701 
 
 q
a
=η I
iθA 
 q
a
 = (0.701)(900)(3.125 
 q
a
 = 1972 W 
 
 q
a
= wc
p
(t
ao
 - t
ai
) 
 w = 25 kg/s 
 c
p
 = 4190 J/kg.K 
 
 1972 = (25)(4190)(t
ao
 - 60) 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 7 of 12 
 
 t
ao
 = 127.8 C - - - Ans. 
 
 
 (c) If qa = 0 
 
 Eq. 20-11 
 
( )( )FrttUI
A
q
aic2c1i
a
∞θ −−αττ== a0
 
 
( )( )( ) ( )( )( )0.932t6.50.90.99000 ai −−= 
 t
ai
 = 144.2 C - - - Ans. 
 
20-7. Two architects have different notions of how to orient windows on the west side of a building in order to be 
most effective from a solar standpoint-summer and winter. The windows are double-glazed. The two design 
are shown in Fig. 20-15. Compute at 40
o
 north latitude the values of I
T
 from Eq. (20-14) for June 21 at 2 and 
6 P.M. and January 21 at 2 P.M. and then evaluate the pros and cons of the two orientations. 
 See Fig. 20-15. 
 
Solution: 
 Eq. 20-14 
 
( )τφ= cosII DNT 
 (a) For notion (a). 
 
 At 40
o
 north latitude, June 21 at 2 P.M. 
 δ = 23.5o 
 H = 2 x 15 = 30
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 30 cos 23.5 + sin 40 sin 23.5 
 β = 59.85 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
cos59.85
30cos23.5sin
sin =φ
 
 
65.91=φ
 
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = 30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 65.91- 60 = 5.91 
 cos θ = cos 59.85 cos 5.91 sin 30 + sin 59.85 cos 30 
 cos θ = 0.99866 
 θ = 87 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 8 of 12 
 
 Fig. 20-6, Double Glazing 
 τ = 0.11 
 
( )τφ= cosII DNT 
 




 β
=
sin
Bexp
A
IDN
 
 A = 1080 W/m
2
 in Mid-summer 
 B = 0.21 in summer 
 
 
( )( )( )( )
sin59.85
0.21exp
0.110.998661080
IT =
 
 I
T
 = 93.06 W/m
2
 
 
 40
o
 north latitude, June 21, 6 P.M. 
 δ = 23.5o 
 H = 6 x 15 = 90
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 90 cos 23.5 + sin 40 sin 23.5 
 β = 14.85 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
cos14.85
90cos23.5sin
sin =φ
 
 
71.58=φ
 
 But Table 4-13, 
 
90>φ
 
 
108.4271.58-180 ==φ
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = 30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 108.42- 60 = 48.42 
 cos θ = cos 14.85 cos 48.42 sin 30 + sin 14.85 cos 30 
 cos θ = 0.542703 
 θ = 57.13 
 
 Fig. 20-6, Double Glazing 
 τ = 0.68 
 
( )τφ= cosII DNT 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 9 of 12 
 




 β
=
sin
Bexp
A
IDN
 
 A = 1080 W/m
2
 in Mid-summer 
 B = 0.21 in summer 
 
 
( )( )( )( )
sin14.85
0.21exp
0.680.5427031080
IT =
 
 I
T
 = 175.65 W/m
2
 
 
 At 40
o
 north latitude, January 21 at 2 P.M. 
 
( )
365
N284360
23.47sin
+
=δ
 
 N = 21 
 
( )
20.16
365
21284360
23.47sin −=
+
=δ
 
 
 δ = -20.16o 
 H = 2 x 15 = 30
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 30 cos (-20.16) + sin 40 sin (-20.16) 
 β = 23.66 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
( )
cos23.66
sin30-20.6cos
sin =φ
 
 
30.83=φ
 
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = 30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 30.83- 60 = -29.17 
 cos θ = cos 23.66 cos (-29.17) sin 30 + sin 23.66 cos 30 
 cos θ = 0.747434 
 θ = 41.63 
 
 Fig. 20-6, Double Glazing 
 τ = 0.76 
 
( )τφ= cosII DNT 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 10 of 12 
 




 β
=
sin
Bexp
A
IDN
 
 A = 1230 W/m
2
 in December and January 
 B = 0.14 in summer 
 
 
( )( )( )( )
sin23.66
0.14exp
0.760.7474341230
IT =
 
 I
T
 = 493 W/m
2
 
 
Then: 
June 21, 2 P.M. I
T
 = 93.06 W/m
2
 
June 21, 6 P.M. I
T
 = 175.65 W/m
2
 
January 21, 2 P.M. I
T
 = 493 W/m
2
 
 
 (b) For notion (b). 
 
 At 40
o
 north latitude, June 21 at 2 P.M. 
 δ = 23.5o 
 H = 2 x 15 = 30
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 30 cos 23.5 + sin 40 sin 23.5 
 β = 59.85 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
cos59.85
30cos23.5sin
sin =φ
 
 
65.91=φ
 
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = -30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 65.91- 60 = 5.91 
 cos θ = cos 59.85 cos 5.91 sin (-30) + sin 59.85 cos (-30) 
 cos θ = 0.499066 
 θ = 60.06 
 
 Fig. 20-6, Double Glazing 
 τ = 0.65 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 11 of 12 
 
( )τφ= cosII DNT 
 




 β
=
sin
Bexp
A
IDN
 
 A = 1080 W/m
2
 in Mid-summer 
 B = 0.21 in summer 
 
 
( )( )( )( )
sin59.85
0.21exp
0.650.4990661080
IT =
 
 I
T
 = 274.8 W/m
2
 
 
 40
o
 north latitude, June 21, 6 P.M. 
 δ = 23.5o 
 H = 6 x 15 = 90
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 90 cos 23.5 + sin 40 sin 23.5 
 β = 14.85 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
cos14.85
90cos23.5sin
sin =φ
 
 
71.58=φ
 
 But Table 4-13, 
 
90>φ
 
 
108.4271.58-180 ==φ
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = -30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 108.42- 60 = 48.42 
 cos θ = cos 14.85 cos 48.42 sin (-30) + sin 14.85 cos (-30) 
 cos θ = -0.0988 
 θ = 95.67 > 90 
Therefore 
 I
T
 = 0.00 W/m
2
 
 
 At 40
o
 north latitude, January 21 at 2 P.M. 
 
( )
365
N284360
23.47sin
+
=δ
 
 N = 21 
CHAPTER 20 - SOLAR ENERGY 
 
 Page 12 of 12 
 
( )
20.16
365
21284360
23.47sin −=
+
=δ
 
 
 δ = -20.16o 
 H = 2 x 15 = 30
o
 
 L = 40
o
 
 
 Eq. 20-3 
 sin β = cos L cos H cos δ + sin L sin δ 
 sin β = cos 40 cos 30 cos (-20.16) + sin 40 sin (-20.16) 
 β = 23.66 
 
 Eq. 20-4 
 β
δ
=φ
cos
sinHcos
sin
 
 
( )
cos23.66
sin30-20.6cos
sin =φ
 
 
30.83=φ
 
 
 Eq. 20-8. 
 
 cos θ = cos β cos γ sin Σ + sin β cos Σ 
 Σ = tilt angle = -30 
 For facing west, Eq. 20-6. 
 
ϕ±φ=γ
 
 
60=ϕ
 
 γ = 30.83- 60 = -29.17 
 cosθ = cos23.66cos(-29.17)sin (-30)+ sin 23.66 cos(-30) 
 cos θ = -0.05238 
 θ = 93 > 90 
Therfore 
 I
T
 = 0.00 W/m
2
 
 
Then: 
June 21, 2 P.M. I
T
 = 274.80 W/m
2
 
June 21, 6 P.M. I
T
 = 0.00 W/m
2
 
January 21, 2 P.M. I
T
 = 0.00 W/m
2
 
 
Ans. 
Design (b) is most effective on June 21 at 2 P.M. but least effective on June 21 at 6 P.M. Design (a) is most effective 
on January 21, at 2 P.M. 
 
- 0 0 0 - 
 
CHAPTER 21 - ACOUSTICS AND NOISE CONTROL 
 Page 1 of 4 
21-1. A tube 1.5 m long has a speaker at one end and a reflecting plug at the other. The frequency of a pure-tone 
generator driving the speaker is to be set so that standing waves will develop in the tube. What frequency is 
required? 
 
Solution: Eq. 21-3. 
 λ = c / f 
 λ = x = 1.5 m 
 c = 344 m/s 
 
 f = c/ λ = (344 m/s) / (1.5 m/s) 
 f = 229 Hz 
 
 At x = 2λ 
 λ = x / 2 = 1.5 m / 2 = 0.75 m 
 
 f = (344 m/s) / (0.75 m) = 458 Hz 
 
 Ans. 229 Hz, 458 Hz, etc., 
 
21-2. The sound power emitted by a certain rocket engine is 10
7
 W, which is radiated uniformly in all directions. 
(a) Calculate the amplitude of the sound pressure fluctuation 10 m removed from the source. 
(b) What percentage is this amplitude of the standard atmospheric pressure? 
 
Solution: 
 (a) 
 
Watts
2c
Ap
E
2
o
ρ
=
 
 
2r4Α pi= 
 ρ
pi
=
2c
pr4
E
2
o
2
 
 r = 10 m 
 c = 344 m/s 
 ρ = 1.18 kg/m3 
 E = 10
7
 W 
 
( )
( )( )
7
2
o
2
10
1.183442
p104
E =
pi
=
 
 p
o
 = 2,542 Pa - - - Ans. 
 
 (b) 
 
 
0.0251
Pa101,325
Pa2,542
Percentage ==
 
 Percentage = 2.51 % - - - Ans. 
 
 
21-3. At a distance of 3 m from a sound source of 100 W that radiates uniformly in all directions what is the SPL 
due to direct radiation from this source? 
 
Solution: 
 Combine Eq. 21-8 and Eq. 21-9. 
CHAPTER 21 - ACOUSTICS AND NOISE CONTROL 
 Page 2 of 4 
 
2
2
rms
r4
E
c
p
pi
=
ρ 
 
2
2
rms
r4
cE
p
pi
ρ
=
 
 E = 100 W 
 c = 344 m/s 
 ρ = 1.18 kg/m3 
 r = 3 m 
 
( )( )( )
( )2
2
rms
4
3441.18100
p
3pi
=
 
 
22
rms Pa359p = 
 p
ref
 = 20 µPa = 20x10-6 Pa 
 
 Eq. 21-11. 
 
( ) 





×
==
−
261020
359
10log
2
ref
2
rms
p
p
log10SPL
 
 SPL = 119.5 dB - - - Ans. 
 
21-4. An octave-band measurement resulted in the following SPL measurements in decibels for the eight octave 
bands listed in Table 21-1: 65.4, 67.3, 71.0, 74.2, 72.6, 70.9, 67.8, and 56.0, respectively. What is the 
expected overall SPL reading? 
 
Solution: 
 
1210−
∑
=
I
log10SPL
 
 Eq. 21-14. 
 
1112
1 SPLIL
10
I
10log ==
−
 
 






=
− 10
SPL
12 1010I
 






+++++++= −∑ 10
56
10
67.8
10
70.9
10
72.6
10
74.2
10
71
10
67.3
10
65.4
12 101010101010101010I
 
 
84,653,020
10
I
12
=
−
∑
 
 
1210−
∑
=
I
log10SPL
 
 
( )84,653,02010logSPL =
 
 SPL = 79.3 dB - - - Ans. 
 
21-5. A room has a ceiling area of 25 m
2
 with acoustic material that has an absorption coefficient of 0.55; the walls 
and floor have a total area of 95 m
2
 with an absorption coefficient of 0.12. A sound source located in the 
center of the room emits a sound power level of 70 dB. What is the SPL at a location 3 m from the source? 
 
Solution: 
 inc
abs
I
I
=α
 
CHAPTER 21 - ACOUSTICS AND NOISE CONTROL 
 Page 3 of 4 
 Eq. 21-17. 
 21
2211
SS
aSaS
+
+
=α
 
 S
1
 = 25 m
2
, α
1
 = 0.55 
 S
2
 = 95 m
2
, α
2
 = 0.12 
 
( )( ) ( )( )
9525
0.12950.5525
+
+
=α
 
 0.2096=α 
 SPL = 70 dB 
 
 Eq. 21-18. 
 
 α−
α
=
1
S
R
 
 S = 25 m
2
 + 95 m
2
 = 120 m
2 
 
( )( ) 2m31.82
0.20961
0.2096120
R =
−
=
 
 Fig. 21-9, at Distance = 3 m 
 SPL - PWl = -8 
 SPL = PWL - 8 
 SPL = 70 - 8 
 SPL = 62 dB - - - Ans. 
 
21-6. In computing the transmission of sound power through a duct, the standard calculation procedure for a 
branch take-off is to assume that the sound power in watts divides in ratio of the areas of the two branches. If 
a PWL of 78 dB exists before the branch, what is the distribution of power in the two branches if the areas of 
the branches (a) are equal and (b) are in a ration of 4:1? 
 
Solution: PWL = 78 dB 
 
 Eq. 21-10. 
 
 o
E
E
10logPWL =
 
 
( ) ( )
63,095,7351010
E
E
10
78
10
PWL
o
===
 
 
 (a) For equal area: 
 
 
( ) 31,547,73563,095,735
2
1
E
E
o
1
==
 
 PWL
1
 = 10log(31,547,868) 
 PWL
1
 = 75 dB - - - Ans. 
 
 (b) For ratio of 4:1: 
 
 
( ) ,619,14563,095,735
5
1
E
E
o
2 12==
 
 
CHAPTER 21 - ACOUSTICS AND NOISE CONTROL 
 Page 4 of 4 
 PWL
2
 = 10log(12,619,145) 
 PWL
2
 = 71 dB - - - Ans. 
 
 
( ) ,476,58863,095,735
5
4
E
E
o
3 50==
 
 
 PWL
3
 = 10log(50,476,588) 
 PWL
3
 = 77 dB - - - Ans. 
 
- 0 0 0 - 
	Chapter2
	Chapter3
	Chapter4
	Chapter5
	Chapter6
	Chapter7
	Chapter8
	Chapter9
	Chapter10
	Chapter11
	Chapter12
	Chapter13
	Chapter14
	Chapter15
	Chapter16
	Chapter17
	Chapter18
	Chapter19
	Chapter20
	Chapter21